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Math 120A — An Introduction to Group Theory
Neil Donaldson
Fall 2015
Text
• An Introduction to Abstract Algebra, John Fraleigh, 7th Ed 2003, Adison–Wesley (optional).
• Also check the library for entries under “Group Theory” and “(Abstract) Algebra”. There have
been a plethora of textbooks written for Undergraduate group theory courses and the first few
chapters of most of them will cover similar ground to the core text.
1
Introduction
Why study Groups?
Here are two possible answers. Firstly groups are incredibly useful. Groups are largely about symmetries and patterns and much of mathematics involves looking for these. For example a square
has symmetries (rotations and reflections): these symmetries form a group. Here are some further
examples of where groups play a role.
Geometry A large part of modern Geometry involves the study of groups — surfaces, polyhedra,
the set of lines in a vector space — these sets have many different groups associated to them
which help the geometer describe and classify them.
Combinatorics When studying collections of objects, groups of permutations (reorderings) of sets
are widely used.
Galois Theory Groups describe symmetries in the roots of polynomial equations.
Chemistry Groups are used to describe the symmetries of molecules and of crystalline substances.
Physics Materials science sees group theory in a similar way to Chemistry, while modern theories
of the nature of the Universe (e.g. string theory) rely heavily on groups.
The second reason to study groups is that they are (relatively) easy — yes really. A group is a set
with just one operation (like the real numbers R with +) — you are already familiar with objects far
more complicated than this (e.g. R with the two operations +, · is a field, (Rn , +, ·) is a vector space)
— so it makes sense to begin the study of abstract mathematics at the level of groups. Since there is
only one operation, the number of options is very limited. Sometimes doing the only thing you can will
lead you to a correct proof!
1
Motivational example
What do an equilateral triangle and an arbitrary collection {1, 2, 3} of 3 objects have in common?
Not much at first glance, but both have symmetries: rotations and reflections of the triangle and
permutations of the set {1, 2, 3}. If one considers compositions of these symmetries or permuations,
we will see that, in a fundamental way, these symmetries are the same.
Permuations of {1, 2, 3} These can be written in cycle notation:1 for example, the cycle (12) represents swapping 1 and 2 and leaving 3 alone. As a function:
(12)(1) = 2,
(12)(2) = 1,
(12)(3) = 3.
Similarly, (123) sends 1 to 2, 2 to 3 and 3 to 1. It is not hard to convince yourself that there are six
distinct permutations of {1, 2, 3}: for brevity, we will use the symbols e, µ1 , µ2 , µ3 , ρ1 , ρ2 .
Identity: leave everything alone Swap two numbers Permute all three
e = ()
µ1 = (23)
ρ1 = (123)
µ2 = (13)
ρ2 = (132)
µ3 = (12)
We can compose these permuations. For instance
µ1 ◦ ρ2 = (23)(132)
maps each of the three numbers as follows:


1 7 → 3 7 → 2
2 7→ 1 7→ 1


3 7→ 2 7→ 3
The result is the same as that obtained by the permuation (12) = µ3 , whence we write
µ1 ◦ ρ2 = µ3 .
The full list of compositions of the symmetries is shown below (left column first, then top row):
◦
e
ρ1
ρ2
µ1
µ2
µ3
1 We
e ρ1 ρ2
e ρ1 ρ2
ρ1 ρ2 e
ρ2 e ρ1
µ1 µ2 µ3
µ2 µ3 µ1
µ3 µ1 µ2
µ1
µ1
µ3
µ2
e
ρ2
ρ1
µ2
µ2
µ1
µ3
ρ1
e
ρ2
µ3
µ3
µ2
µ1
ρ2
ρ1
e
will return to all this notation later.
2
The Equilateral Triangle What does this have to do with
a triangle? If we label the vertices of an equilateral trian3
gle 1,2,3, then the above permutations correspond to symmetries of the triangle: ρ1 and ρ2 are rotations, while each
µi performs a reflection in the altitude through the vertex i.
ρ1
ρ2
The two sets of symmetries apply to different objects, but
µ2
µ1
their interactions are seen to be identical.
One might also get intuition about the permutations of
{1, 2, 3} by viewing them as symmetries of a triangle. In particular, there is a qualitative difference between the rotations
ρ1 , ρ2 and the reflections µ1 , µ2 , µ3 . The fact that composition
of two reflections makes a rotation is clear in the triangle set2
1
µ3
ting. That the composition of two 2-cycles makes a 3-cycle is
not so clear.
Group theory is about ideas like the above: the symmetries and patterns associated to an object
are often seen to be more important and useful than the object itself and can lead to unexpected
connections. In group notation we have shown that the symmetric group on 3 letters S3 (permutations of {1, 2, 3}) and the Dihedral group of order 6 D3 (the symmetries of the equilateral triangle) are
isomorphic.2 This will be written S3 ∼
= D3 .
2
Binary relations
Definition 2.1. A binary relation ∗ on a set X is a map ∗ : X × X → X which is defined for all pairs of
elements in X. We say that X is closed under ∗.
Examples
1. Addition + and multiplication · are binary relations on any of the usual sets of numbers: R, C,
Q, Z, R+ , Q+ , Z+ (and many more).
2. The sets R, Q, C, Z with ∗ being subtraction.
3. Any vector space (e.g. Rn or Cn ) has addition as a binary relation.
4. R3 with cross/vector product ×.
5. {Functions f : V → V }, where V is any set and ∗ = ◦ (composition of functions).
6. C with ∗ the Hermitian product: z ∗ w = zw.
7. N = Z+ with ∗ the least common multiple: e.g. 4 ∗ 5 = 20.
8. N with ∗ the highest common prime factor: e.g. 65 ∗ 20 = 5.
9. The set A = {0, 1, 2, 3} with a ∗ b = ab mod 4 (i.e. the remainder upon dividing ab by 4).
2 We will explain both examples more fully later and terms such as ‘isomorphic’. At the moment it is enough to say that
isomorphic means ‘the same in a specific way’.
3
Remarks There is no requirement for a binary relation on X to have an image that is all of X (i.e.
X ∗ X need not be all of X; see example 8 where a ∗ b is always prime). The only thing that matters is
that a ∗ b is always an element of X. Here are two non-examples of binary relations:
1. (Q, ∗) with a ∗ b = a/b is not a binary relation since a ∗ 0 is undefined for all a.
2. Let M = {2 × 2 matrices with determinant 7} with A ∗ B = AB being matrix multiplication.
Since
det( AB) = det A det B
it follows that det( A ∗ B) = 49 and so A ∗ B 6∈ M. ∗ is not a binary relation on M.
Relation Tables
Binary relations on finite sets can be written in (multiplication) tables. Example 9 above can be
written
0
0
0
0
0
∗
0
1
2
3
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
where we find a ∗ b by taking a in the left column and b along the top.
Definition 2.2. If Y ( X is a proper subset and X has a binary relation ∗ then we can restrict ∗ to Y.
If ∗ restricted to Y is a binary relation on Y, then we say that Y is closed under ∗. Otherwise said, Y is
closed under ∗ if
∀y1 , y2 ∈ Y, y1 ∗ y2 ∈ Y.
Examples
1. The negative integers Z− are closed under + but not under ·.
2. Let Y ⊂ Z be the set of integers whose remainder is 1 when divided by 3: i.e.
Y = {x ∈ Z : x ≡ 1
Is Y closed under +? Or ·?
mod 3} = {1, 4, 7, 10, 13, . . . , −2, −5, −8, . . .}.
• Under +: 1 and 4 are both in Y, yet 1 + 4 = 5 6∈ Y, so Y is not closed.
• Under ·: (1 + 3n) · (1 + 3m) = 1 + 3n + 3m + 9mn = 1 + 3(m + n + 3mn) has remainder
1, whence Y is closed.
Associativity and Commutativity
Definition 2.3. A binary relation ∗ on X is associative if ∀ a, b, c ∈ X, we have a ∗ (b ∗ c) = ( a ∗ b) ∗ c.
If ∗ is associative then writing a1 ∗ a2 ∗ · · · ∗ an is unambiguous. Examples 1,3,5,7,8,9 in the first
list in this section are associative.
Definition 2.4. A binary relation ∗ is commutative on X if a ∗ b = b ∗ a, ∀ a, b ∈ X.
Original examples 1,3,7,8,9 are commutative.
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Remarks Some algebraists use + for any commutative binary relation. While this reminds us of the
common addition of numbers (which is commutative) it can be confusing: do not assume that every
time you see + in a book that it means addition!
The multiplication table of a commutative binary relation necessarily has diagonal symmetry. For example our earlier table is symmetric about the principal diagonal &. Conversely, non-commutative
relations must have non-symmetric tables.
∗
0
1
2
3
0
0
0
0
0
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
∗
a
b
c
d
A commutative relation (ex 9)
a
d
a
a
c
b
c
b
c
c
c
d
a
a
a
d
c
d
b
c
(2.1)
A non-commutative relation
In fact, the second table is not associative either, for ( a ∗ b) ∗ c = c ∗ c = a while a ∗ (b ∗ c) = a ∗ a = d.
Proving associativty can be extremely tedious. A standard approach is to somehow view a binary
relation as being a composition of functions and appealing to the following result.
Theorem 2.5. Let V be any set. Composition of functions V → V is associative.
If V has at least 2 elements, then composition is non-commutative.
Proof. For the first claim we must see that ( f ◦ g) ◦ h = f ◦ ( g ◦ h), ∀ functions f , g, h : V → V. To
see that two functions are the same, it is enough to see that they map all elements x ∈ V to the same
place. Now
(( f ◦ g) ◦ h)( x ) = ( f ◦ g)(h( x )) = f ( g(h( x ))) and,
( f ◦ ( g ◦ h))( x ) = f (( g ◦ h)( x )) = f ( g(h( x ))).
Thus ◦ is associative.
To see that ◦ is non-commutative we have to exhibit a counter-example for any V with at least two
elements. Let a 6= b be two elements in V and define f , g : V → V by f ( x ) = a, g( x ) = b, ∀ x ∈ V.
Then,
f ◦ g : x 7→ f (b) = a,
g ◦ f : x 7→ g( a) = b.
◦ is therefore non-commutative.
Remarks You may think it is enough to choose a specific space (say V = R) in which to exhibit a
counter-example, but the proof really requires the abstractness.
There are many large sets of functions that do commute: for example in a vector space V the set of
scalings f λ : V → V : x 7→ λx where λ ∈ R form an infinite commuting set.
Composition of functions between arbitrary sets is actually associative (the proof is almost identical),
but composition only gives a binary relation when the functions are from a set to itself.
Corollary 2.6. Matrix multiplication of square matrices is associative.
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Proof. We will provide three proofs of this fact — choose your favourite.
Using Theorem 2.5 You may have proved in a linear algebra course that matrix multiplication corresponds to linear transformations of a vector space: i.e. square matrices can be viewed as
functions V → V where multiplication of matrices corresponds to composition of linear transformations. Armed with this knowledge, Theorem 2.5 gives the result straightaway.
Directly Suppose that A, B, C are n × n matrices. Recall that if A has entries aij (ith row, jth column),
etc., then AB has ikth entry
( AB)ik =
n
∑ aij bjk .
j =1
Therefore the il th entry of ( AB)C is,
!
(( AB)C )il =
n
∑
k =1
n
∑ aij bjk
j =1
ckl =
n
∑ (aij bjk )ckl .
j,k =1
Similarly
( A( BC ))il =
n
n
∑ aij ∑ bjk ckl
j =1
k =1
n
!
=
∑
j,k =1
aij (b jk ckl ).
These are equal by the associativity of (R, ·).
The Summation Convention If you are lucky enough to understand the Einstein Summation Convention, observe that the direct proof can be written:
( AB)C = ( aij b jk )ckl = aij (b jk ckl ) = A( BC ),
where we have used associativity of multiplication of real numbers for the middle step.
Matrix multiplication is similarly non-commutative: for example A =
commute.
01
00
and B =
00
10
don’t
We will often ignore associativity and just assume it is true. As you can see above, proofs of
associativity often involve a messy calculation. When thought about in the correct way, it is often
easy to view many binary operations as functional composition, whence Theorem 2.5 is all we need.
Identities
An identity for a binary relation on a set X is an element of X which behaves very simply with respect
to the binary relation: it does nothing at all!
Definition 2.7. Let ∗ be a binary relation on X. e ∈ X is an identity for ∗ if
∀ x ∈ X, e ∗ x = x ∗ e = x.
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Examples and non-examples
1. 0 is an identity for addition on any set of numbers including zero.
2. 1 is an identity for multiplication on any set of numbers including one.
3. There is no identity for the binary relation of the cross product × on R3 . Suppose that an
identity e ∈ R3 existed. Then, for example,
e×i = i
However, e × i must be perpendicular to i. A contradiction!
4. The identity function on any set is the function with formula e( x ) = x. If f is any function on
this set, then e ◦ f = f ◦ e = f .
Theorem 2.8. Identities are unique. Otherwise said, if a binary relation ∗ on X has an identity e ∈ X, then
there are no others.
Proof. Suppose that e, ê ∈ X are identities. Then
and
e ∗ ê = ê,
since e is an identity,
e ∗ ê = e,
since ê is an identity.
In particular e = ê.
The proof illustrates an earlier point: often there is only one thing you can do, so why not do it?
To show that something is unique, assume there are two and show that they are equal. Since the only
structure we have is that of the binary relation ∗, the only thing we can do with two elements is to
combine them using ∗. Once this is done, the proof is almost immediate. Have a little faith, and often
the only thing you can do is the correct thing to do!
The roots of unity
An important collection of examples of a binary relations are given by the nth roots of unity under
multiplication.
Definition 2.9. Let n ∈ N. The nth roots of unity are the (complex) solutions to the equation zn = 1.
Proposition 2.10. If we label ζ n = e
are precisely the values
2πi
n
(usually just ζ if the context is known), then the nth roots of unity
1, ζ, ζ 2 , · · · , ζ n−1 .
Proof. Take the modulus of both sides of zn = 1 to obtain |z|n = 1. Since |z| is a positive real number
(it certainly can’t be zero!), the only solution to this is |z| = 1. Thus all such z lie on the unit circle in
C and must be of the form eiθ . Now compute:
1 = zn = (eiθ )n = einθ ⇐⇒ nθ = 2πk, for some integer k.
Thus θ =
e
2πi
n k
2πk
n
and so
2πi k
= en
= ζk.
7
The nth roots are equally spaced around the unit circle, forming the corners of a regular n-gon
centered at the origin, with one corner at 1.
e
2πi
3
e
i
−1
e
6πi
7
e
8πi
7
1
e
4πi
3
−i
i
e
2πi
7
−1
1
e
Cube roots of unity
4πi
7
−i
10πi
e
12πi
7
7
Seventh roots of unity
One of the nice properties of the nth roots of unity is that they are closed under multiplication
ζ k · ζ l = ζ k+l
Try to distance yourself from thinking about the what the roots are, and instead simply on how they
interact. For example, the cube roots under multiplication have the following relation table:
·
1
ζ
ζ2
1 ζ ζ2
1 ζ ζ2
ζ ζ2 1
ζ2 1 ζ
Compare this to another binary relation, that of addition modulo 3 on the set {0, 1, 2}. Its table is
+3
1
1
2
0
0
1
2
1
1
2
0
2
2
0
1
The entries in the two tables are different, but the structure, the pattern is the same. We want to
encapsulate this mathematically. One way to do this is to define a function φ : {0, 1, 2} → {1, ζ, ζ 2 }
between the two sets which is bijective and preserves the structure on both sides. An example would be
φ( x ) = ζ x
All the entries in the first table are simply φ of the corresponding entry in the second table. Indeed,
by the exponential laws (ζ x+y = ζ x ζ y ), we immediately have
∀ x, y ∈ {0, 1, 2},
φ( x + y) = φ( x )φ(y)
so that φ transfers the structure of addition modulo 3 on {0, 1, 2} to multiplication on {1, ζ, ζ 2 }. We
will call φ an isomorphism of binary structures.
8
3
Morphisms
On its own, a morphism is simply a function which preserves some structure. The type of structure
depends on the context. In this course, all morphisms will have one of the prefices homo- or iso-.3
Definition 3.1. Let X, Y be sets with binary relations ∗ X , ∗Y . A map φ : ( X, ∗ X ) → (Y, ∗Y ) is a
homomorphism (of binary structures) if
∀ a, b ∈ X,
φ ( a ∗ X b ) = φ ( a ) ∗Y φ ( b ) .
If the binary structures are known to the reader, one typically just writes φ : X → Y.
Recall the following definitions.
Definition 3.2. Let X, Y be sets. A function (or morphism) f : X → Y is said to be;
1–1 (injective) if f ( x ) = f (y) =⇒ x = y for all x, y ∈ X,
onto (surjective) if f ( X ) = Y.
Otherwise said: f is 1–1 iff each element in the image of f comes from exactly one element of X;
f is onto iff everything in Y is in the image of f .
f is a bijection if it is both 1–1 and onto.
Definition 3.3. A homomorphism φ : ( X, ∗ X ) → (Y, ∗Y ) is an isomorphism4 if φ : X → Y is 1–1 and
φ
∼ Y or X ∼
onto. We write X ∼
= Y or occasionally φ : X ∼
= Y, φ : X −→
= Y. Some books (e.g. Fraleigh)
−
use X ' Y.
Remarks Recall the motivational example where we discussed the symmetries of the equilateral
triangle D3 and the permutations of {1, 2, 3}. With both sets of transformations labelled the way they
are we have an isomorphism S3 ∼
= D3 , with binary operation of composition on each side.
Pull-backs Given a bijection φ : X → Y and a binary relation ? on Y, you can always define a
relation ∗ on X by pulling-back ? to X:
x1 ∗ x2 := φ−1 (φ( x1 ) ? φ( x2 ))
In fact, this makes sense when φ is merely injective. There is a similar notion of push-forward.
Examples
1. The map φ : R+ → R+ : x 7→ x2 is a bijection. Suppose we want φ to be an isomorphism
φ : (R+ , ∗) → (R+ , +). Then we must define ∗ by
q
x ∗ y := φ−1 (φ( x ) + φ(y)) = φ−1 ( x2 + y2 ) = x2 + y2 .
3 Homo
4 ‘Iso’
means ‘similar’ and iso means ‘same.’
means ‘same’.
9
2. Now suppose that we want φ to be an isomprohism φ : (R+ , +) → (R+ , ?). This time we must
have φ satisfying
φ( x ) ? φ(y) = φ( x + y) =⇒ x2 ? y2 = ( x + y)2
√
√
It follows that the required binary relation ? is X ? Y = ( X + Y )2 .
Showing Isomorphicity
When asked to demonstrate that two structures are isomorphic (e.g. that ( X, ∗) ∼
= (Y, ?)) you should
follow the steps given below:
• Define a function φ : X → Y,
• Check that φ is a homomorphism: φ( x ∗ y) = φ( x ) ? φ(y),
• Check φ is 1–1: φ( x ) = φ(y) =⇒ x = y,
• Check φ is onto: ∀z ∈ Y, ∃ x ∈ X such that z = φ( x ).
You can perform the final three steps in any order you like: sometimes it is easier to think of a bijection
and check that it is a hommorphism, other times it is easier to use the homomorphism property to
help you decide on a function, then check that you have a bijection.
Examples
1. 2Z ∼
= 3Z with the binary relation of addition on both sides.
Define φ Let φ : 2Z → 3Z : 2n 7→ 3n, i.e. φ( x ) = 23 x.
Homomorphism φ( x + y) = 32 ( x + y) = 23 x + 32 y = φ( x ) + φ(y) X
1–1 φ( x ) = φ(y) =⇒
3
2x
= 23 y =⇒ x = y, so φ is 1–1 X
Onto An arbitrary element of 3Z is z = 3m where m ∈ Z. But 3m = φ(2m), so φ is onto X
φ is therefore an isomorphism φ : 2Z ∼
= 3Z.
2. The bijection φ( x ) = 32 x is also an isomorphism 2Z ∼
= 3Z with respect to the binary relations
1
1
x ∗ y = 2 xy on 2Z and x ? y = 3 xy on 3Z.
Showing non-isomorphicity
This is tricky in general: you can’t try every map X → Y (except for very small sets X, Y) in order to
check that there are no isomorphisms, so you have to be a little more cunning and consider various
properties that are preserved by isomorphism.
Definition 3.4. A structural property of ( X, ∗) is a property which is preserved under isomorphisms:
i.e. if φ : ( X, ∗) → (Y, ∗0 ) is an isomorphism then ( X, ∗) and (Y, ∗0 ) have the same structural properties.
The following are examples of structural properties: suppose φ : ( X, ∗) → (Y, ∗0 ) is an isomorphism throughout.
10
• If X ∼
= Y, then the elements of X, Y are bijectively paired ( x, φ( x )), whence the size (cardinality)
of X, Y must be the same. This is true even for infinite sets: recall that |Z| = |Q| = ℵ0 2ℵ0 =
|R|, so there are no isomorphisms between Q and R, no matter what binary relations you have.
• Commutativity & Associativity: If X or Y has one of these properties, then so must the other.
A set with a commutative (or associative) relation is never isomorphic to a set with a noncommutative (or non-associative) relation.
• Identity: if ( X, ∗) has an identity element e (e ∗ x = x ∗ e = x, ∀ x ∈ X) then, under an isomorphism φ, Y has an identity φ(e).
• Any ‘special’ elements in X must correspond to special elements in Y: for example idempotents,
if x ∗ x = x in X, then under an isomorphism φ we must have φ( x ) ? φ( x ) = φ( x ) in Y. Thus if
idempotents exist in X, but don’t in Y, there can be no isomorphism.
• Solutions of equations more generally: if a ∗ x = b has a solution x ∈ X, then, in the presence
of an isomorphism, the corresponding equation φ( a) ? φ( x ) = φ(b) must have a solution in Y.
There are many other things that must be preserved by an isomorphism, far too many to list. The
more structure a set and its binary operators get, the more properties there are to consider. Disproving isomorphisms is somewhat of an Art, as you will sometimes have to think very hard or have a
flash of inspiration in order to think of the correct structural property. There are also many concepts
that are not necessarily preserved by an isomorphism: the type of element (a number, a matrix, etc.),
the type of binary operation (e.g. addition versus multiplication).
Examples
1. The two binary relations defined by the multiplication tables in (2.1) are non-isomorphic because the first is commutative while the second is not.
2. (R3 , +) (R3 , ×) (cross product), because + is commutative and associative, while × is neither.
3. (N0 , +) (N, +) since N0 contains an identity element 0 while N does not.
4. (Q, +) (Q+ , ·) since y2 = 2 has no solution in Q, but the corresponding equation x + x = c
has a solution in Q for any c.
4
Groups
Definition 4.1. A Group is a set G together with an operation · : G × G → G which together satisfy
the following axioms:
Closure G is closed under · (i.e. · is a binary relation on G),
Associativity · is associative,
Identity ∃ identity e ∈ G such that e · g = g · e = g, ∀ g ∈ G,
Inverse Every element has an inverse: ∀ g ∈ G, ∃ g−1 ∈ G such that g · g−1 = g−1 · g = e.
Definition 4.2. A group G is Abelian if the group operation is commutative.5
5 In
honor of the Norwegian mathematician Niels Abel, one of the godfathers of pure mathematics.
11
Notation and Conventions Although a group is often written as a pair ( G, ·), it is conventional to
just write G if there is no ambiguity about the operation.
Abstract groups are commonly written multiplicatively, as in the above definition. The group operation can simply be written as juxtaposition. I.e. g · h = gh.
In a multiplicative group the power notation is a convenient short-hand: i.e.
an = |a ·{z
· · }a
and
a − n = ( a −1 ) n = ( a n ) −1
ntimes
Exponents follow the usual additive law an am = am+n . We also write a0 = e.
Abstract Abelian groups are sometimes written additively, the operation being denoted +. In such
cases we use minus (−) for the inverse. Multiplicaton then becomes a shorthand for repeated additions: e.g. 3x = x + x + x.
If a group is non-abstract—you have an explicit set and operation—you should use the correct symbol
for the operation!
Examples
1. The sets C, R, Q, Z are Abelian groups under addition:
Check All are closed by assumption X
Addition is associative X
0 is the additive identity: 0 + x = x + 0 = x X
The inverse of x is − x X
2. The non-zero elements in C, R, Q form Abelian groups under multiplication.
Check Closure: x, y 6= 0 =⇒ xy 6= 0 X
Multiplication of numbers is associative X
1 ∈ Q ⊂ R ⊂ C is the multiplicative identity: 1 · x = x · 1 = x X
The inverse of x is x −1 = 1x , which is in C, R, Q if n is X
These groups are often denoted R× , C× , Q× , etc.
3. Vector spaces: any vector space is an Abelian group under addition. Note that this includes the
set of m × n matrices under addition as ( Mm×n (R), +) = (Rm×n , +).
4. Let P be a regular n-gon. Then P has symmetries (rotations and reflections). The set of symmetries forms a non-Abelian group under composition. These are the Dihedral Groups Dn . We will
think more about these later.
5. Permutation groups: the symmetric group Sn is the group of permutaions of n letters, typically
of the set {1, 2, . . . , n} we will also consider these later.
6. Matrix groups: several sets of matrices form groups under multiplication. For example, the
Orthogonal Group O(n) of transformations of Rn which preserve lengths and angles between
vectors. The identity matrix I, with 1s down the main diagonal and 0s elsewhere is the suitablynamed identity for these groups.
There are many, many more examples!
12
Things that are not groups
There are a number of things that can go wrong when trying to make a set into a group, even if you
overcome the biggest hurdle and start with an associative operation. Mostly people forget to check
closure and inverses.
• The set of all real numbers R (as opposed to R× = R \ {0}) does not form a group under
multiplication. (R, ·) is closed, associative and has the identity element 1, but the single element
0 has no inverse: @r ∈ R such that 0 · r = r · 0 = 1.
The same is true for the other multiplicative sets: (Q, ·) and (C, ·) are not groups because of the
non-invertibility of 0.
• If V is a set with at least two elements, then the set of functions f : V → V does not form
a group under composition. We have closure, associativity (Theorem 2.5) and an identity map
f : x 7→ x, but in general we have no inverses. For example the map f ( x ) = constant has
no inverse. However if the space V is itself a group, then the set of maps f : X → V from
any set into V forms a group under the operation of V on values: for example { f : X → Rn }
forms a group under addition on Rn ; ( f + g)( x ) = f ( x ) + g( x ). In this way, for example, the
continuous functions form a group under addition.
• The set of functions f : R → R is not a group under multiplication of values. Any function f
which has a zero at any point x has no inverse, for f (1x) is undefined.
Basic theorems for groups
Theorem 4.3. In a group G, identities and inverses are unique.
Proof. We already proved that identities are unique for any binary structure in Theorem 2.8.
Now suppose that g has two inverses h1 6= h2 . Then gh1 = gh2 (= e). Thus
h1 ( gh1 ) = h1 ( gh2 )
(by associativity)
=⇒ (h1 g)h1 = (h1 g)h2
=⇒ eh1 = eh2
=⇒ h1 = h2 .
(since h1 is an inverse of g)
Again we have a contradiction.
Notice how we used associativity in the proof. Do you think that there might exist an algebraic
structure which is non-associative in which inverses are non-unique?
Corollary 4.4 (Cancellation laws & Inverses). In any group G we have:
(a) gg1 = gg2 =⇒ g1 = g2 .
(b) g1 g = g2 g =⇒ g1 = g2 .
(c) ( gh)−1 = h−1 g−1 .
13
Proof. The first two statements are immediate - just multiply on the left (resp. right) by g−1 . For part
(c), note that
h−1 g−1 ( gh) = h−1 ( g−1 g)h = h−1 eh = h−1 h = e.
Thus h−1 g−1 is an inverse of gh. But inverses are unique by Theorem 4.3.
Definition 4.5. Two groups G, H are isomorphic if the group operations on the sets G, H are isomorphic binary structures: We write G ∼
= H.
Written multiplicatively, G ∼
= H ⇐⇒ ∃ a bijective φ : G → H which is also a homomorphism
φ ( g1 g2 ) = φ ( g1 ) φ ( g2 )
Group tables for finite groups
Just as for binary relations on finite sets, it can be useful to write a group in a multiplication table: for
groups, these are known as Cayley tables.6 There are a few rules for Cayley tables however which do
not apply to general binary relations.
In the row and column of the identity e, everything remains unchanged. You can always start by
writing down a group table with the first row and column filled out.
·
e
a
b
c
e a b c
e a b c
a
b
c
Complete the table with any entries you like and you automatically have the closure and identity
axioms satisfied.
Lemma 4.6 (Magic square lemma). In a Cayley table, each element appears exactly once in each row and
column.
Proof. Suppose an element x appears twice in row a. Then ∃b 6= c such that x = ab = ac. But
Corollary 4.4 implies b = c: a contradiction. Considering column a is similar. Since no element can
appear twice, we must have all elements appearing exactly once.
Any table satisfying the identity and magic square rules defines a binary relation with identity
and unique inverses: i.e. the Closure, Identity & Inverse axioms are satisfied. The only thing left to
check is associativity. Unfortunately this is very messy to see directly from the table, and diminishes
the use of tables as a good method of defining groups. Another disadvantage of Cayley tables is that
large isomorphic tables can appear to be very different.
6 After
the British mathematician Arthur Cayley, one of the founders of group theory.
14
Group tables for low order groups
Definition 4.7. The order of a group G is the cardinality of G. Groups with infinite cardinalities are
usually referred to as infinite groups.
By following the magic square approach, we can easily construct all possible Cayley tables for small
order groups. Indeed it is easy to see that there is only one group each (up to isomorphism) of orders
1, 2 and 3.
·
e
e
e
Order 1
·
e
a
b
e a
e a
a e
·
e
a
Order 2
e a b
e a b
a b e
b e a
Order 3
When represented abstractly, as above, these groups are often denoted C1 , C2 and C3 , where C
stands for cyclic. Do we know of any explicit sets and binary relations which have these group
structures?
Theorem 4.8. The set of remainders {0, 1, 2, . . . , n − 1} forms an Abelian group under addition modulo n:
a +n b := a + b (mod n). We label this group Zn .
Proof. Let’s check the axioms:
Closure a +n b is a remainder modulo n. X
Associativity a +n (b +n c) ≡ a + (b +n c) ≡ a + b + c (mod n). This is symmetric in a, b, c and
therefore +n is associative. X
Identity 0 is the additive identity. X
Inverse − a ≡ n − a (mod n). Thus n − a is the inverse of a. X
(Zn , +n ) is therefore a group. +n is moreover commutative (since + is) and so Zn is Abelian.
Corollary 4.9. There exists a group of every finite order. I.e. for every positive integer n there exists at least
one group with n elements (namely (Zn , +n )).
Now return to the Cayley tables above. If we write out the table for (Z3 , +3 ), we obtain
+3
0
1
2
0
0
1
2
1
1
2
0
2
2
0
1
It should be obvious that φ :
0
1
2
7→
e
a
b
is an isomorphism φ : (Z3 , +3 ) ∼
= C3 .
Strictly speaking Zn = Z ∼ = {[0], [1], [2], . . . , [n − 1]} is the set of equivalence classes in Z under the equivalence relation a ∼ b ⇐⇒ a ≡ b (mod n). The excess of brackets makes this unwieldy,
so we typically treat Zn as above. The equivalence relation description will become very important
towards the end of the course.
15
5
Subgroups
Definition 5.1. Let G be a group and H ⊆ G a subset. H is a subgroup of G if H is a group in its own
right, with respect to the same group operation.
A subgroup H is a proper subgroup if H 6= G.
{e} is the trivial subgroup. All other subgroups are non-trivial.
We write H ≤ G, or H < G to stress that H is a proper subgroup.
The definition of subgroup contains several points. In particular if H is a subgroup of G then the
following are true:
• H ⊆ G as sets.
• The group operation on G restricts to a binary operation on H (H is closed).
• The identity e ∈ G must lie in H.
• H is closed under inverses, h−1 ∈ H ∀h ∈ H. A priori we only have that h−1 ∈ G for h ∈ H, so
this really is a condition.
Examples
1. (Z, +) < (Q, +) < (R, +) < (C, +).
2. (Q× , ·) < (R× , ·) < (C× , ·).
3. {e} ≤ G and G ≤ G for any G.
4. (Rn , +) < (Cn , +).
5. (Rm , +) ≤ (Rn , +) if m ≤ n.
6. (S1 , ·) < (C× , ·) (recall S1 = {z ∈ C : |z| = 1} is the unit circle).
7. (C (R), +) < (C1 (R), +) (all differentiable functions are continuous).
You don’t have to check all the axioms of a group to see that a subset is a subgroup, indeed you
only have to check the closure and inverse axioms, as the following result makes clear.
Theorem 5.2. A subset H of G is a subgroup iff H is closed under the group operation and inverses. That is:
(
∀h1 , h2 ∈ H, h1 h2 ∈ H
H ≤ G ⇐⇒
∀h ∈ H, h−1 ∈ H
Proof. (⇒) If H is a subgroup, then it is a group, and thus satisfies the closure and inverse axioms.
(⇐) Since H is a subset of G, it is automatic that the group operation of G is associative on H. Our
assumption is that H also satisfies the closure and inverse axioms. It remains therefore only to check
the identity axiom.
By assumption, if h ∈ H we have
eG = hh−1 ∈ H,
since inverses and products remain in H. The identity eG of G is therefore in H, and so H is a
subgroup.
You should show directly that the above examples are subgroups using the Theorem.
16
Subgroups of low order groups
Recall the groups C1 , C2 , C3 under modulo addition. It is straightforward to check that the only
subgroups these groups have are the identity group {e} ∼
= C1 and themselves. Their subgroup
structures are not very interesting! When we reach order 4, more interesting things start to happen.
It is an exercise to see that, up to isomorphism, there are two groups of order 4, with the following
Cayley tables:
c
∗ e a b c
∗ e a b c
e e a b c
e e a b c
a a b c e
a a e c b
b
a
b b c e a
b b c e a
c c e a b
c c b a e
The cyclic group C4
The Klein 4-group V
Representation of V
We already have a concrete interpretation of the cyclic group, namely the remainders Z4 under
addition modulo 4. What about the Klein 4-group?7 This can be represented as the symmetries of a
regular rhombus (or rectangle): choose one of a, b, c to be rotation by 180◦ and the remaining terms
to be reflections.
Observe that V C4 since the equation x2 = e has 4 solutions in V while having only 2 solutions
in C4 — recall that the existence of solutions to such an equation are a structural property of a group:
if V and C4 were isomorphic then both must have the same number of solutions.
Now consider subgroups. C4 has three subgroups: C4 , {e, b} ∼
= C2 and {e} ∼
= C1 . You should
check that there are no other subgroups. For example, if we tried to include a in a subgroup, then we
would also have to include a2 = b and a3 = c, whence we obtain the entire group C4 .
V by contrast has five subgroups: V, {e, a}, {e, b}, {e, c}, and {e} (the middle subgroups are all isomorphic to C2 ). Each two-element set is a subgroup because all elements of V are their own inverse.
If we tried to make a group out of {e, a, b}, then we would be forced to include ab = c: the subset
{e, a, b} is not closed and therefore not a subgroup.
Subgroup diagrams
It can be helpful to draw subgroup relations. A line in a diagram means that the lower group is a
subgroup of the upper.
C1
C2
C3
C4
V
{e, b} ∼
= C2
{e} ∼
= C1
C1
{e, a} ∼
= C2
C1
{e, b} ∼
= C2
{e, c} ∼
= C2
C1
It should be obvious that the number and relationship of subgroups are structural properties.
This gives another proof that V and C4 are non-isomorphic.
7 Named
for Felix Klein. V refers to the German vier, meaning ‘four.
17
Geometric Subgroup Proofs
Sometimes an argument for being a subgroup is much easier to make geometrically. Recall that the
set of symmetries of any physical object is a group under composition. Suppose that you can arrange
two objects in such a way that every symmetry of the first is also a symmetry of the second. Then
immediately the first group is a subgroup of the second.
For example, a regular hexagon can have an equilateral triangle drawn inside it in two distinct
ways.
µ4
µ3
µ2
µ5
ρ4
µ0 ρ2
µ1
ρ4
ρ2
Here the six reflections of the hexagon are labelled µ0 , . . . , µ5 and the rotations e, ρ1 , . . . , ρ5 ; for
example, ρ2 rotates the hexagon two steps, or 120◦ , counter-clockwise. Each of the six symmetries of
the equilateral triangle is also a symmetry of the hexagon. It follows that the symmetry group D3 of
the triangle is a subgroup of that of the hexagon D6 in two different ways:
D3 ∼
= {e, ρ2 , ρ4 , µ0 , µ2 , µ4 } ∼
= {e, ρ2 , ρ4 , µ1 , µ3 , µ5 }
Can you see how to demonstrate that the Klein 4-group is a subgroup of D6 ? In how many distinct ways can this be done?
What about showing that the symmetry group of a cube is a subgroup of that of a sphere?
Matrix groups
There are many groups of matrices under multiplication. These are of particular interest to geometers, because they are often defined with an intention of preserving certain geometric properties: e.g.
volume, length, angle, etc.
Definition 5.3. The general linear group GLn (R) is the set of invertible linear transformations of an
n-dimensional real vector space under composition. Since all such vector spaces in the presence of
a basis may be viewed as Rn , this group may be represented by the set of invertible n × n matrices
under multiplication.
GLn (R) = { A ∈ Mn (R) : det A 6= 0}
This is certainly a group:
Closure If A, B ∈ GLn (R), then det( AB) = det A · det B 6= 0, whence AB ∈ GLn (R).
Associativity Holds by Corollary 2.6.
Identity In = diag(1, 1, . . . , 1) is the multiplicative identity.
Inverse If A ∈ GLn (R), then the matrix inverse satisfies det( A−1 ) =
non-zero. Thus A−1 ∈ GLn (R).
18
1
det A
which is defined and
Special linear group
The first property that we might want to consider is measure (i.e. area in 2D, volume in 3D, etc.). You
should have seen the following in a linear algebra class.
Theorem 5.4. Suppose that P is a parallelogram spanned by
the vectors u, v. Then its signed area is
u1 v1
Area( P) = |u| |v| sin θu,v = det
u2 v2
θ is measured counter-clockwise from u to v, whence the
signed area is positive iff 0 < θ < π.
v
P
θ
u
Now suppose that A is a 2 × 2 matrix. Let us compute what A does to the parallelogram P. It is
easy to check that the result is a new parallelogram spanned by the vectors Au and Av. We can now
compute its signed area:
( Au)1 ( Av)1
u v
Area( A( P)) = det
= det A 1 1
= det A · Area( P)
( Au)2 ( Av)2
u2 v2
If we want the new parallelogram A( P) to have the same signed area as P, then we must restrict to
matrices with determinant 1. This set of matrices will be called the special linear group SL2 (R).
Definition 5.5. The special linear group SLn (R) is the group under multiplication of n × n matrices
with determinant 1.
We can check that this is a subgroup by appealing to Theorem 5.2: suppose that A, B ∈ SLn (R),
then
det( AB) = det A det B = 1
and
det( A−1 ) =
1
=1
det A
whence SLn (R) is a subset of GLn (R) closed under multiplication and inverses.
The interpretation of SL3 (R) is similar to that of SL2 (R).
The signed volume of a parallelepiped in R3 spanned by
the vectors u, v, w is the scalar triple product
Volume = [u, v, w] = (u × v) · w
If A is a 3 × 3 matrix, then the parallelepiped spanned by Au, Av, Aw will have volume
[ Au, Av, Aw] = det( A)[u, v, w]
SL3 (R) is therefore the group of signed8 volume-preserving linear transformations of R3 .
8 u × v points perpendicular to the plane spanned by the two vectors. A parallelepiped has positive signed-volume if
w points to the same side of this plane as u × v. If u points out of the page and v into the page, then the picure denotes a
parallelepiped with positive signed volume. This convention is essentially the right-hand rule.
More generally, transformations with det > 0 are said to preserve orientation. SLn (R) is then the set of orientation- and
(hyper-)volume-preserving linear transformations of Rn . Quite the mouthful.
19
The Orthogonal group
The orthogonal group is the set of matrices which preserve lengths of vectors and the angles between
them. Recall that both of these concepts can be described in terms of the scalar/dot product:
Scalar product: u · v = u T v
√
Length: |u| = u · u
Angle: cos θ =
u·v
|u| |v|
If we want a matrix A to preserve angles between and lengths of vectors, it is enough to have the
matrix preserve the value of the scalar product of any two vectors. That is, we require
∀u, v,
( Au) · ( Av) = u · v
Using the transpose represenation, this is the same as
u T v = ( Au)T ( Av) = u T A T Av
Definition 5.6. The orthogonal group is the group under multiplication of matrices
O n ( R ) = { A ∈ Mn ( R ) : A T A = I }
where I is the n × n identity matrix diag(1, . . . , 1).
The special orthogonal group SOn (R) = On (R) ∩ SLn (R) is the subset of determinant 1 orthogonal
matrices.
Taking determinants of the property A T A = I gives det( A) det( A T ) = det( A)2 = 1
det( A) = ±1. Thus SOn (R) consists of exactly half the orthogonal group.
=⇒
Interpretations When n = 2 the group SO2 (R) consists of the rotations around the origin (det = 1)
while O2 (R) also includes the set of reflections across lines through the origin (det = −1).
SO3 (R) comprises all rotations around the origin. O3 (R) also includes reflections across any plane
through the origin.
Other Matrix Groups: non-examinable
Pseudo-orthogonal groups The scalar product definition of On (R) can be extended. Suppose you
put a non-positive-definite scalar product on Rn , for example

1 0
0
0
0 −1 0
0 

(u, v) := uT 
0 0 −1 0  v
0 0
0 −1

on R4 . The group which preserves this inner product is the pseudo-orthogonal group O(1, 3). This
example is much more than abstract mathematical nonsense: in Physics this is the Lorentz Group,
which is critical to the study of relativity.
20
Complex matrix groups All of the above examples can be constructed with complex entries, yielding the groups GLn (C), On (C), etc.
The Unitary group The unitary group is constructed similarly to the orthogonal group. This time
we take the Hermitian product on Cn : hu, vi = u∗ v = u T v. The unitary group is the set of complex
matrices which preserve this product:
T
Un = { A ∈ GLn (C) : A A = I }.
The special unitary group SUn is the set of determinant 1 unitary matrices, that is SUn = Un ∩ SLn (C).
U1 = (S1 , ·) is the unit circle under multiplication. U1 is the set of 1 × 1 complex matrices (i.e.
numbers z ∈ C) which satisfy zz = 1. Thus |z| = 1 =⇒ z ∈ S1 . Conversely all numbers on the circle
have the form eiθ for some θ ∈ [0, 2π ) from which we see that eiθ eiθ = e−iθ eiθ = 1.
A discrete matrix group Not all matrix groups must be continuous like the above. The set
SLn (Z) = { A ∈ SLn (R) : all entries are integers}
is a group. It is clearly a subset of SLn (R) and it is easy to see that the product of two matrices with
integer entries has integer entries, so it remains to check inverses. Recall from linear algebra the
formula
A −1 =
1
adj( A)T ,
det A
where adj( A) is the adjoint of A, formed by taking positive and negative multiples of determinant
minors of A. Observe that if A ∈ SLn (Z), then adj has integer entries and so therefore has its transpose. Since det( A) = 1 it follows that the inverse not only has determinant 1, but has integer entries.
SLn (Z) is therefore a group.
A subgroup diagram for matrix groups For reference, the subgroup relations between the various
groups are summarised in the following table.
GLn (C)
GLn (R)
On (C )
SLn (C)
Un
SLn (R)
On (R )
SOn (C)
SUn
SLn (Z)
SOn (R)
There are many more matrix groups out there!
21
6
Cyclic groups
Cyclic groups are a very basic class of groups: we have already seen some such as (Zn , +n ) (Theorem
4.8). Cyclic groups are relatively easy to work with since their complete structure is easy to describe.
The overall approach in this section is to define and classify all cyclic groups and understand their
subgroup structure.
Definition 6.1. Let G be a group (written multiplicatively) and g ∈ G. The cyclic subgroup of G
generated by g is the subgroup
h g i = { g n : n ∈ Z}.
The order of an element g ∈ G is the order |h gi| of the subgroup generated by g. G is a cyclic group iff
∃ g ∈ G such that G = h gi.
We now have two concepts of order. The order of a group is the cardinality of the group, while the
order of an element is the cardinality of the cyclic group generated by that element. It follows that
cyclic groups are the only groups containing elements having the same order as that of the group.
Examples
Integers The integers Z form a cyclic group under addition. Z is generated by either 1 or −1. Note
that this group is written additively, so that, for example, the subgroup generated by 2 is the
group of even numbers under addition:
h2i = {2m : m ∈ Z} = 2Z
Modular Addition For each n ∈ N, the group of remainders Zn under addition modulo n is a cyclic
group. It is also written additively, and is generated by 1. Typically Zn is also generated by
several other elements. For example, Z4 = {0, 1, 2, 3} is generated by both 1 and 3:
h1i = {1, 2, 3, 0}
h3i = {3, 2, 1, 0}
Roots of Unity For each n ∈ N, the nth roots of unity Un = {1, ζ n , . . . , ζ nn−1 } form a cyclic group
under multiplication. This group is generated by ζ n , amongst others. The generators of Un are
termed the primitive nth roots of unity.
We shall see shortly that every cyclic group is isomorphic to either the integers or the modular integers. Regardless, we will still write abstract cyclic groups multiplicatively. Let us start the
classification of cyclic groups with the following Lemma.
Lemma 6.2. Every cyclic group is Abelian.
Proof. Let G = h gi. Then any two elements of G can be written gk , gl for some k, l ∈ Z. But then
gk gl = gk+l = gl +k = gl gk ,
and so G is Abelian.
Note that the converse is false: the Klein 4-group V is Abelian but not cyclic.
22
Before going any further, we need to distinguish between finite and infinite groups.
Lemma 6.3. Let G = h gi be cyclic. Then the set S = {m ∈ N : gm = e} is non-empty iff G is a finite group.
Proof. Suppose that S is empty and suppose that gk = gl for some k > l. But then gk−l = e is a
contradiction. It follows that the elements gk are all distinct, and so G is an infinite group.
Now suppose that S is non-empty and contains an element n. Then
∀λ ∈ Z,
gk+λn = gk
whence there are at most n elements in G.
Finite Cyclic Groups Now we prove a fundamental theorem which directly ties finite cyclic groups
to modular arithmetic.
Theorem 6.4. Suppose that G = h gi has finite order n. Then gk = e ⇐⇒ k = qn for some q ∈ Z. In
particular,
order( G ) = min{m ∈ N : gm = e}
Moreover, G ∼
= (Zn , + n ).
Proof. Let n = min{m ∈ N : gm = e}. This exists by the well-ordering principle. Certainly we have
gqn = e for all q ∈ Z.
Now assume that gk = e. Apply the Division algorithm to k and n: thus k = qn + r, where q, r ∈ Z
and 0 ≤ r < n. Then
e = gk = gqn+r = ( gn )q gr = gr .
Since r < n and n is the minimal positive power returning the identity, this forces r = 0, and consequently k = qn.
Now observe that
gk = gl ⇐⇒ gk−l = e ⇐⇒ k − l ≡ 0
(mod n) ⇐⇒ k ≡ l
(mod n).
(∗)
It follows that the group G = | g| contains precisely n elements.
It remains to define an isomorphism φ : Zn → G. Let
φ(k ) = gk
Well-definition There is something to prove here: remember that k ∈ Zn is an equivalence class. We
must therefore check that if k ≡ l (mod n), then φ(k ) = φ(l ). But this is precisely (∗) above. X
1–1 φ(k) = φ(l ) =⇒ gk = gl =⇒ k ≡ l (mod n), also by (∗). X
Onto Since every element of G has the form gk = φ(k ), it follows that φ is onto. X
Homomorphism φ(k +n l ) = gk+n l = gk+l = gk gl = φ(k )φ(l ). X
Observe that the proofs of well-definition and 1–1 are essentially the converses of each other!
23
Example The group of 7th roots of unity (U7 , ·) is isomorphic to (Z7 , +7 ) under the isomorphism
φ : Z7 → U7 : k 7→ ζ 7k
Infinite Cyclic Groups Now we repeat our analysis for infinite cyclic groups.
Theorem 6.5. If G is an infinite cyclic group, then G ∼
= (Z, +).
Proof. Let G = h gi be an infinite cyclic group. Define φ : Z → G the same way as before by φ(k ) = gk .
The argument that this is an isomorphism is almost identical to the previous proof: the primary
difference is that we do not need to check well-definition.
1–1 Suppose that φ(k ) = φ(l ) for some integers k > l. Then gk = gl =⇒ gk−l = e. By Lemma 6.3,
G must be finite. Contradiction. X
Onto Since every element of G has the form gk = φ(k ), it follows that φ is onto. X
Homomorphism φ(k + l ) = gk+l = gk gl = φ(k )φ(l ). X
An example and a non-example (Z, +) has many subgroups. As we sill see below, all except the
trivial subgroup {0} are isomorphic to the integers themselves. For example, 5Z = h5i, the group of
multiples of 5, is isomorphic to the integers via
φ : (Z, +) ∼
= (5Z, +) : z 7→ 5z
The real numbers R form an infinte group under addition. This cannot be cyclic because its cardinality
2ℵ0 is larger than the cardinality ℵ0 of the integers.
What is the cyclic group of order n? Strictly speaking, there is no such thing as the cyclic group
Cn of order n. If an algebraist says that a particular group is Cn , they simply mean that the group
is cyclic and has order n. Such is therefore isomorphic to (Zn , +n ) and to any other cyclic group of
order n.
Particular examples of cyclic groups of order n are often referred to as presentations of Cn . The fact
that Cn is somewhat nebulous turns out to be convenient. For example: in some texts you may find
Z6 referred to as the cyclic group of order 6. What, then, should we call the following subgroup of
(Z12 , +12 )?
h2i = {2, 4, 6, 8, 10, 0}
Certaintly h2i is isomorphic to Z6 via φ : 2z 7→ z. It would be erroneous to say that h2i equals Z6 ,
since Z6 = {0, 1, 2, 3, 4, 5} has different elements, namely the remainders (or equivalences classes of
integers) modulo 6.
However, if we allow C6 to refer to any cyclic group of order 6, it seems more reasonable to write
h2i = C6 . This is merely a convenient way of saying that the subgroup h2i is cyclic of order 6.
We will follow this convention: G = Cn means that G is cyclic of order n.
24
Subgroups of cyclic groups
We can very straightforwardly classify all the subgroups of a cyclic group.
Theorem 6.6. All subgroups of a cyclic group are cyclic.
Proof. Let G = h gi and let H ≤ G. Since all elements of H are in G, there must exist9 a smallest
natural number n such that gn ∈ H. We claim that H = h gn i.
For this let gm ∈ H be a general element of H. But then we may divide m by n so that there exist
unique integers q, r ∈ Z, 0 ≤ r < n satisfying
m = qn + r.
Therefore
gm = gqn+r = ( gn )q gr ∈ H.
But gn ∈ H =⇒ ( gn )q ∈ H and so
gr = ( gn )−q gm ∈ H.
However n is the smallest natural number so that gn ∈ H: since 0 ≤ r < n we are forced to conclude
that r = 0. In summary gm = ( gn )q ∈ h gn i and we are done.
Subgroups of infinite cyclic groups If G is an infinite cyclic group, then any subgroup is itself
cyclic and thus generated by some element. Since all infinite cyclic groups are isomorphic to (Z, +)
it is easies to think about this group. Clearly the subgroup generated by n ∈ Z is
hni = nZ = {nz : z ∈ Z}
namely the multiples of n under addition. If n = 0 this is the trivial group, but if n 6= 0 then nZ is
infinite and thus isomorphic to Z itself. Indeed
φ : Z → nZ : z 7→ nz
is an isomorphism. It should be clear that nZ ≤ mZ ⇐⇒ n | m.
Having completely described the subgroup of infinite cyclic groups, we now turn our attention
to finite cyclic groups.
Theorem 6.7. Let G = h gi have order n. The subgroup H ≤ G generated by the element h = gs is the cyclic
group C nd where d = gcd(s, n).
Proof. We know that hhi is a cyclic group of finite order (since is a subgroup of the finite group G). It
is enough therefore to check that hk = e ⇐⇒ k ≡ 0 (mod nd ).
(⇐) If k = α nd , then
n
s
s
hk = ( gs )k = ( gs )α d = ( gn )α d = eα d = e
9 Well-ordering
again. . .
25
since ds ∈ Z.
(⇒) Conversely,
hk = e =⇒ ( gs )k = e ⇐⇒ gsk = e
=⇒ sk = αn for some α ∈ Z
s
n
=⇒ k = α
d
d
n
=⇒ k ≡ 0
mod
d
Therefore hhi = C nd .
((∗) in Theorem 6.4)
(since d divides s and n)
(since gcd ds , nd = 1)
Corollary 6.8. A cyclic group of order n has precisely one subgroup C nd for each divisor d of n.
In particular, if G = h gi has order n, then h gs i = gt ⇐⇒ gcd(s, n) = gcd(t, n).
We will prove the Theorem for the group Zn , since the notation is simpler.
Proof. Let d be a divisor of n. Then by the Theorem, hdi = C nd = C gcd(nd,n) .
Conversely, suppose that gcd( a, n) = d. Then,
• a = σd for some σ ∈ Z, so that a ∈ hdi and so h ai ≤ hdi.
• By the Euclidean Algorithm, ∃λ, µ ∈ Z such that λa + µn = d, and so λa ≡ d (mod n). Thus
d ∈ h ai, from which hdi ≤ h ai.
In conclusion: if gcd( a, n) = d then h ai = hdi = C nd .
Examples
1. Z8 = {0, 1, 2, 3, 4, 5, 6, 7} is generated by 1, 3, 5, 7, e.g.
h5i = {5, 2, 7, 4, 1, 6, 3, 0} = Z8 .
The subgroup generated by 6 is then
h6i = {6, 4, 2, 0}
which has order 4 =
8
gcd(6,8)
in accordance with the Theorem.
2. Here we find all the subgroups of Z30 . Note that 30 = 2 · 3 · 5. We list all the elements of Z30
according to their greatest common divisor with 30, and then the subgroup that each generates.
According to the above results, each subgroup in the right column is generated by any of the
numbers in the left column.
x
gcd( x, 30) 30/ gcd( x, 30) subgroup generated
0(30)
0(30)
1
C1
15
15
2
C2
10, 20
10
3
C3
6, 12, 18, 24
6
5
C5
5, 25
5
6
C6
3, 9, 21, 27
3
10
C10
2, 4, 8, 14, 16, 22, 26, 28
2
15
C15
1, 7, 11, 13, 17, 19, 23, 29
1
30
Z30
26
Here is the subgroup diagram for Z30 with the obvious generator chosen for each subgroup.
Z30 = h1i
C15 = h2i
C10 = h3i
C6 = h5i
C5 = h6i
C3 = h10i
C2 = h15i
C1 = h0i
7
Generating sets — non-examinable
Definition 7.1. If X ⊆ G is a subset of a group G then the subgroup of G generated by X is the
subgroup created by making all possible combinations of elements and inverses of elements in X.
The subgroup generated by X will be written
hx ∈ Xi
G is finitely generated if there exists a finite subset of G which generates G.
The subgroup of G generated by X really is a subgroup: it is certainly a subset, so we need only
check that it is closed under multiplication and inverses. However, the definition says that we keep
throwing things into the subgroup so that it satisfies precisely these conditions!
Examples
1. (Z, +) = h2, 3i. The inverse of 2 = −2 must lie in h2, 3i. But then 3 + (−2) = 1 ∈ h2, 3i. Since
1 generates Z, we are done.
2. In general if m, n ∈ Z then the subgroup hm, ni = {λm + µn : λ, µ ∈ Z} is hdi = dZ where
d = gcd(m, n). You will return to this concept in a Number Theory course.
3. (Q, +) = n1 : n ∈ Z+ . (Q, +) is not finitely generated: there exists no finite subset which
p
generates Q. To see this, consider the subgroup generated by some finite set X = { qii }in=1 . Any
fraction created by adding and subtracting elements of this set must have a denominator which
is a product of one or more of the qi ’s, perhaps with some cancelling with the numerator. It is
therefore impossible to generate the number 1q from X if q is any number relatively prime to all
the qi .
If this seems a little tricky, consider that if X = { 12 , 13 }, then
3m + 2n
k
: m, n ∈ Z =
:k∈Z
(for reasons similar to example 2)
hx ∈ Xi =
6
6
Clearly this subgroup does not contain 15 .
27
8
Permutation groups
In the earliest conceptions of group theory, all groups were considered permutation groups. Essentially
a group was a collection of ways in which one could rearrange some set or object: for example, each
‘rearrangement’ of an equilateral triangle corresponds to one of its symmetries. It was Arthur Cayley,
of Cayley-table fame, who formulated the group axioms we use now. Importantly, he also proved
what is now known as Cayley’s Theorem: that the old definition and the new are in fact identical.
So what, first, is a permutation?
Definition 8.1. A permutation of a set A is a bijection ϕ : A → A.
Theorem 8.2. The set of permutations S A of any set A forms a group under composition.
Proof. We check the axioms:
Closure If ϕ, ψ are bijective then so is the composition ϕ ◦ ψ. X
Associativity Permutations are functions, the composition of which we know to be associative (Theorem 2.5). X
Identity The identity function ι A maps all elements of A to themselves: id A : x 7→ x. This is certainly
bijective. X
Inverse If ϕ is a permutation then it is a bijection, whence the function ϕ−1 exists and is also a
bijection. X
Definition 8.3. The symmetric group on n-letters Sn is the group of permutations of any set A of n
elements. Typically we choose A = {1, 2, . . . , n}.
Proposition 8.4. Sn has n! elements (unlike Cn or Zn where the subscript is the order of the group).
Proof. This is just counting. If σ ∈ Sn , then there are n choices for the value of σ (1). Choosing one
leaves n − 1 choices for σ(2) (since σ is a bijection). Iterating this argument we get ... 2 choices for
σ(n − 1) and only 1 possibility for σ(n). We therefore have n(n − 1) · · · 2 · 1 = n! possibilities in
total.
To describe a group as a permutation group simply means that each element of the group is being
viewed as a permutation of some set. As the following Theorem shows, all groups are permutation
groups, although sadly not in a particularly useful way!
Theorem 8.5 (Cayley’s Theorem). Every group G is isomorphic to a group of permutations.
Proof. For each element a ∈ G, let ρ a : G → G be the function ρ a : g 7→ ag (i.e. left multiplication by
a). We make two claims:
1. Each ρ a is a permutation of G.
2. ({ρ a : a ∈ G }, ◦) forms a group isomorphic to G.
The first claim is straightforward. ρ a−1 is the inverse function to ρ a :
∀ g ∈ G,
(ρ a−1 ◦ ρ a )( g) = a−1 ag = g = idG ( g)
whence each ρ a is a bijection.
Now define a map φ : G → {ρ a } by φ( a) = ρ a . We claim this is an isomorphism:
28
1–1 φ( a) = φ(b) =⇒ ρ a = ρb =⇒ ∀ g ∈ G, ag = bg =⇒ a = b. X
onto Certainly every permutation ρ a is in the image of φ. X
Homomorphism If a, b ∈ G, then
φ( a) ◦ φ(b) : g 7→ ρ a (ρb ( g)) = abg = ρ ab ( g)
from which φ( ab) = φ( a) ◦ φ(b).
Note that Cayley’s Theorem is not saying that every group is isomorphic to some Sn . It is saying
that every group is isomorphic to some subgroup of some Sn .
Two notations for permutations
Standard notation Suppose that σ ∈ S4 is the following map
 
 
1
3
2
1

 
σ:
3 7→ 4 , i.e. σ(1) = 3, σ (2) = 1, σ(3) = 4, σ(4) = 2.
4
2
We could then write
1 2 3 4
σ=
,
3 1 4 2
where you read down columns to find where σ maps an element in the top row. Composition
is read in the usual way for functions, do the right permutation first. Thus if
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
ρ=
then σρ =
=
.
1 4 3 2
3 1 4 2
1 4 3 2
3 2 4 1
σ can also be viewed as acting on the vector (1, 2, 3, 4)T , which suggests a matrix method for
encoding elements of Sn . For example, our permutation σ may be written


0 0 1 0
1 0 0 0

σ=
0 0 0 1
0 1 0 0
as a permutation matrix: multiplying on the left by such a matrix permutes the entries of vectors.
Definition 8.6. A permutation matrix is an n × n matrix with exactly one entry of 1 in each row and
column and the remaining entries 0.
Indeed we have proved:
Proposition 8.7. The set of n × n permutation matrices forms a group under multiplication which is isomorphic to Sn .
By Cayley’s Theorem, every finite group of permutations is isomorphic to a group of matrices.
29
Cycle notation Our example permutation can be more compactly written as σ = (1 3 4 2). We
read from left to right, looping back to 1 at the end, each entry telling us where the previous is
mapped to. Thus (1 3 4 2) maps
1 7→ 3 7→ 4 7→ 2 7→ 1.
We have shorter cycles if some of the elements are fixed, for example in our two notations
1 2 3 4 5
∈ S5 .
(1 3) =
3 2 1 4 5
Juxtaposition is used for composition:
1 2 3 4
1 2 3 4
1 2 3 4
=
= (1 3 4).
(1 3 4 2)(2 4) =
3 1 4 2
1 4 3 2
3 2 4 1
Remember that multiplication of cycles is really composition of functions: thus although each
cycle is read from left to right when determining how it acts on elements of {1, 2, . . . , n}, we
multiply cycles by considering the rightmost cycle first. For example, in S5 ,


1 7→ 3





2 7 → 3 7 → 5
(1 3 5 4)(2 3 4) : 3 7→ 4 7→ 1



4 7→ 2




5 7→ 4
=⇒ (1 3 5 4)(2 3 4) = (1 3)(2 5 4)
(†)
This notation can be used to describe non-symmetric groups: e.g. the Klein 4-group V can be
written in terms of cycles if we label the corners of a rhombus
2
3
1
V = {e, (1 3), (2 4), (1 3)(2 4)}
4
Definition 8.8. Suppose that k ≤ n. A k-cycle in Sn is an element ( a1 a2 · · · ak ).
Two cycles ( a1 · · · ak ) and (b1 · · · bl ) are disjoint if no element appears in both cycles: that is, if
{ a1 , . . . , ak } ∩ {b1 , . . . , bl } = ∅.
The identity element is the only 0-cycle. It is sometimes written (), if not otherwise denoted by e.
As the example (†) illustrates, when computing the product of several cycles, the result will typically
be a product of disjoint cycles. This will prove very useful in the next section when we discuss orbits.
30
Subgroups relations between symmetric groups It is easy to see that Sm ≤ Sn ⇐⇒ m ≤ n. For
example, fix the final n − m elements of {1, . . . , n} so that
Sm = {σ ∈ Sn : σ(i ) = i, ∀i > m}.
In fact Sm is a subgroup of Sn in precisely (mn ) different ways: each copy of Sm comes from fixing
n − m elements of {1, . . . , n} and there are (n−nm) = (mn ) ways of choosing these fixed elements.
Dihedral groups
Definition 8.9. The dihedral group Dn is the group of symmetries of the regular n-gon.10
Group? Now that we have defined permutation groups it is very easy to see that the dihedral
groups are indeed groups. Observe that a symmetry of an n-gon can be viewed as a permutation of
the corners of the n-gon for which ‘neighbourliness’ is preserved. This is similar to how we viewed
the Klein 4-group above.
For example, if we label the corners of the regular hexagon 1 through
6 then we see that D6 is the set of σ ∈ S6 such that σ(1) is always next
to σ(6) and σ(2), etc.
Clearly this says that Dn ⊆ S6 .
To see that D6 is a subgroup of S6 , we need only note that the composition of two neighbour-preserving transforms must also preserves
neighbours, as does the inverse of such a map.
Elements of Dn
3
2
4
1
5
6
The regular n-gon has 2n distinct symmetries and so | Dn | = 2n. These consist of:
n rotations For each j = 0, . . . , n − 1, let ρ j be rotation counter-clockwise by
n reflections Let µ j be reflection across the line making angle
you put one of the corners of the n-gon on the x-axis!).11
πj
n
2πj
n
radians.
with the positive x-axis (make sure
Remarks Some authors write D2n instead of Dn precisely because | Dn | = 2n: in this course, Dn will
always mean the symmetries of the n-gon.
Every dihedral group Dn is a subgroup of the orthogonal group O2 (R). The correspondence is:




2πj
2πj
2πj
− sin 2πj
sin
cos
cos n
n 
n n 


µj =
ρj =
2πj
2πj
2πj
sin n
cos n
sin n
− cos 2πj
n
It is a good exercise to convince yourself that these matrices really do correspond to the rotations and
reflections claimed. In particular multiply any two of them together and see what you get...
10 Regular
polygon with n sides.
even-sided polygons these are often labelled differently, and split into two subsets of n2 reflections each. The
reflections µi are those which move all the corners of the n-gon, while δi refers to a reflection across a diagonal. We will see
this in our treatment of D4 below.
11 For
31
In what follows we consider the two simplest dihedral groups and investigate their subgroup
structure.
D3 is the group of symmetries of the equilateral triangle. See the introduction for the multiplication
table. Here we write all the elements in permutation notation and exhibit all the subgroups, their
generators and the subgroup diagram.
2
Element
Rotations
ρ0
1
Reflections
3
ρ0 : the identity
ρ1 : rotate counter-clockwise by 2π
3 radians
radians
ρ2 : rotate clockwise by 2π
3
µi : reflect in altitude through i
ρ1
ρ2
µ1
µ2
µ3
Standard
notation
1 2 3
1 2 3
1 2 3
2 3 1
1 2 3
3 1 2
1 2 3
1 3 2
1 2 3
3 2 1
1 2 3
2 1 3
Cyclic notation
e = ()
(123)
(132)
(23)
(13)
(12)
It should be immediately obvious that all the permutations of {1, 2, 3} are elements of D3 , and so
∼
D3 = S3 . Now we consider all of the subgroups of D3 and its subgroup diagram.
Subgroup Isomorph
{ ρ0 }
C1
{ ρ0 , µ1 }
C2
{ ρ0 , µ2 }
C2
{ ρ0 , µ3 }
C2
{ ρ0 , ρ1 , ρ2 }
C3
D3
S3
D3
Generating sets
{ ρ0 }
{ µ1 }
{ µ2 }
{ µ3 }
{ρ1 }, or {ρ2 }
any pair {ρi , µ j } where
i = 1, 2 and j = 1, 2, 3
C3
C2
C2
C2
C1
How can we be certain that there are no other subgroups of D3 ? A careful consideration of generating sets should convince you. For example, suppose that a subgroup contains two reflections:
WLOG suppose these are µ1 , µ2 . We compute the subgroup generated by {µ1 , µ2 }.
It must include
µ1 µ2 = ρ1 ,
ρ21 = ρ2
µ1 ρ1 = µ3
and thus the entire group.
32
D4 is the group of symmetries of the square. It consists of four rotations and four reflections: the
notation δj for reflection across a diagonal is used here, rather than calling all reflections µ j .
Element
2
3
Rotations
ρ0
1
ρ1
ρ2
ρ3
µ1
4
Reflections
ρ0 : the identity
ρ1 : rotate counter-clockwise by π2 radians
ρ2 : rotate counter-clockwise by π radians
ρ2 : rotate counter-clockwise by 3π
2 radians
µi : reflect across midpoints of sides
δi : reflect across diagonals
µ2
δ1
δ2
Standard
notation
1 2 3 4
1 2 3 4
1 2 3 4
2 3 4 1
1 2 3 4
3 4 1 2
1 2 3 4
4 1 2 3
1 2 3 4
2 1 4 3
1 2 3 4
4 3 2 1
1 2 3 4
1 4 3 2
1 2 3 4
3 2 1 4
Cyclic notation
e = ()
(1234)
(13)(24)
(1432)
(12)(34)
(14)(23)
(24)
(13)
All the subgroups are summarised in the following table. In particular, note that D4 S4 : the
latter has many more elements!
Subgroup
Isomorph
{ ρ0 }
C1
{ ρ0 , µ i }
C2
{ρ0 , δi }
C2
{ ρ0 , ρ2 }
C2
{ ρ0 , ρ1 , ρ2 , ρ3 }
C4
{ ρ0 , µ1 , µ2 , ρ2 }
V
{ρ0 , δ1 , δ2 , ρ2 }
V
D4
−
Generating sets
{ ρ0 }
{µi } for each i
{δi } for each i
{ ρ2 }
{ρ1 } or {ρ3 }
{µ1 , µ2 }, {µ1 , ρ2 } or {µ2 , ρ2 }
{δ1 , δ2 }, {δ1 , ρ2 } or {δ2 , ρ2 }
any pair {ρi , µ j } or {ρi , δj }
where i = 1, 3 and j = 1, 2 or
any pair {µk , δl } where k, l = 1, 2
D4
C2
V
C4
V
C2
C2
C2
C2
C1
In the subgroup diagram, the middle C2 is {ρ0 , ρ2 } while the two copies on each side contain
either side the reflections δi or µi .
Subgroup relations between dihedral groups
Dm ≤ Dn ⇐⇒ m | n. Recall the discussion of geometric proofs: join every (n/m)th vertex of a
regular n-gon to get a regular m-gon. Every symmetry of the m-gon is then a symmetry of the n-gon.
33
Reminder: Equivalence relations
Recall the following discussion from previous classes. Much of the rest of the course depends on this
critical concept!
Definition. An equivalence relation ∼ on a set X is a binary condition12 which satisfies:
Reflexivity ∀ x ∈ X, x ∼ x
Symmetry x ∼ y =⇒ y ∼ x
Transitivity x ∼ y and y ∼ z =⇒ x ∼ z
The set [ x ] = {y ∈ X : y ∼ x } is the equivalence class of x.
The quotient of X by ∼ is the set of equivalence classes, and is written X ∼ .
We may therefore write x ∈ [ x ] ∈ X ∼ . I.e. x is an element of the equivalence class [ x ], which in
turn is an element of the set of equivalence classes.
The important theorem regarding equivalence classes states that they are essentially the same as
partitions of X.
1. If ∼ is an equivalence relation on X, then the equivalence classes of ∼ partition13 X.
Theorem.
2. If X =
S
i∈ I
Xi is a partition of X, then ∼ defined on X by
x ∼ y ⇐⇒ ∃ Xi such that x, y ∈ Xi ,
is an equivalence relation.
Proof.
1. Suppose we are given an equivalence relation ∼. Observe first that x ∼ x =⇒ x ∈ [ x ],
so that every element is in some equivalence class. Now suppose x ∼ y. We want to show that
[ x ] = [y]. We do this in two stages:
(a) Let a ∈ [ x ]. Then a ∼ x. By transitivity, a ∼ x and x ∼ y =⇒ a ∼ y. Therefore a ∈ [y] and
we have [ x ] ⊆ [y].
(b) Now let a ∈ [y]. Then a ∼ y. But now by symmetry we have y ∼ x and so transitivity says
a ∼ y and y ∼ x =⇒ a ∼ x. Thus a ∈ [ x ] and so [y] ⊆ [ x ].
It follows that [ x ] = [y] and we’re done.
2. The converse is straightforward: x ∼ y ⇐⇒ x, y in the same Xi is easily seen to satisfy the
reflexive, symmetric and transitive properties.
12 Either
x ∼ y ‘x is related to y’ or x y ‘x is not related to y’. This is not a binary relation: x ∼ y is either true or false,
not an element of the set X.
13 I.e. every element of X is in exactly one equivalence class. Algebraically, X = S X where the subsets X are disjoint:
i
i
i∈ I
Xi ∩ X j 6= ∅ =⇒ Xi = X j .
34
9
Orbits
In this section we continue the idea of a group being a set of permutations. In particular, we will see
how any element σ ∈ Sn partitions the set {1, 2, . . . , n}. This concept will be generalized later when
we consider group actions.
Definition 9.1. The orbit of σ ∈ Sn containing j ∈ {1, 2, . . . , n} is the set
orb j (σ) = {σk ( j) : k ∈ Z} ⊆ {1, 2, . . . , n}
Remarks Observe that orbσk ( j) (σ) = orb j (σ) for any k ∈ Z.
Be careful: each orbit is a subset of the set {1, 2, . . . , n}, not of the group Sn .
Examples If σ ∈ Sn is written in cycle notation (recall Definition 8.8) using disjoint cycles, then the
cycles are the orbits! For example
Orbits of (134) ∈ S5 are {1, 3, 4}, {2}, {5},
Orbits of (12)(45) ∈ S5 are {1, 2}, {3}, {4, 5}.
The same does not hold if the cycles are not disjoint. For example, you may check that σ = (13)(234) ∈
S4 maps
1 7→ 3 7→ 4 7→ 2 7→ 1
whence there is only one orbit: orb j (σ) = {1, 2, 3, 4} for any j. In fact, σ = (1234), from which the
orbit is obvious.
Given that disjoint cycle notation is so useful for reading off orbits, it is a natural question to ask
if any permutation σ can be written as a product of disjoint cycles. The answer, of course, is yes, with
the disjoint cycles turning out to be precisely the orbits of σ!
Theorem 9.2. The orbits of any σ ∈ Sn partition X = {1, 2, . . . , n}.
Proof. Define ∼ on X = {1, 2, . . . , n} by
x ∼ y ⇐⇒ y ∈ orbx (σ)
We claim that ∼ is an equivalence relation.
Reflexivity x ∼ x since x = σ0 ( x ). X
Symmetry x ∼ y =⇒ y = σk ( x ) for some k ∈ Z. But then x = σ−k (y) =⇒ y ∼ x. X
Transitivity Suppose that x ∼ y and y ∼ z. Then y = σk ( x ) and z = σl (y) for some k, l ∈ Z. But
then z = σk+l ( x ) and so x ∼ z. X
The equivalence classes of ∼ are clearly the orbits of σ, which therefore partition X.
Corollary 9.3. Every permutation can be written as a product of disjoint cycles.
35
Proof. Write out each of the orbits of σ ∈ Sn in order, placing each orbit in parentheses ( ). Since the
orbits of σ partition X = {1, 2 . . . , n} the cycles obtained are disjoint.
More concretely,
orb1 (σ) = {1, σ(1), σ2 (1), . . .}
If this orbit is the entirity of X, then we are finished. Otherwise, let
x1 = min{ x : x 6∈ orb1 (σ)}
and construct its orbit:
orbx1 (σ) = { x1 , σ( x1 ), σ2 ( x1 ), . . .}
This orbit must be disjoint with orb1 (σ). Now repeat. It is immediate from the construction that
σ = 1 σ (1) σ 2 (1) · · · x1 σ ( x1 ) σ 2 ( x1 ) · · · · · · · · ·
If you mechanically follow the algorithm for multiplying cycles (see the previous section) you
will automatically end up with a product of disjoint cycles:
Examples
1. (13)(234)(1432) = (123).
2. (13)(24)(12)(34) = (14)(23).
Note that disjoint cycles commute! E.g. (14)(23) = (23)(14).
Now that we are able to write any permutation as a product of disjoint cycles, we are able to
compute much more easily. For example:
Theorem 9.4. The order of a permutation σ is the least common multiple of the lengths of its disjoint cycles.
Proof. Write σ as a product of disjoint cycles σ = σ1 · · · σm . Since disjoint cycles commute, it it immediate that
σn = σ1n · · · σmn
Moreover, since the terms σjn permute disjoint sets, it follows that
σn = e ⇐⇒ ∀ j, σjn = e
A k-cycle clearly has order k (the least positive integer l such that ( a1 · · · ak )l = e). If the orbits of σ
have lengths α j ∈ Z+ respectively, it follows that
σjn = e ⇐⇒ α j | n
Thus n must be a multiple of α j for all j. The least such n is clearly lcm{α j }.
36
Example The order of σ = (145)(3627)(89) ∈ S9 is lcm(3, 4, 2) = 12.
We can easily calculate σ3465 for the above σ. Since 3465 = 12 · 288 + 9 we have
σ3465 = (σ12 )288 σ9 = σ9 = (145)9 (3627)9 (89)9 = (3627)(89),
since (145), (3627) and (89) have orders 3, 4 and 2 respectively.
Transpositions
Instead of breaking a permutation σ into disjoint cycles, we can consider a permutation as being
constructed from only the simplest bijections.
Definition 9.5. A 2-cycle ( a1 a2 ) is also known as a transposition, since it swaps two elements of
{1, 2, . . . , n} and leaves the rest untouched.
Theorem 9.6. Every σ ∈ Sn is the product of transpositions.
Proof. There are many, many ways to write out a single permutation as a product of transpositions.
One method is to first write σ as a product of disjoint cycles, then write each cycle as follows:
( a1 · · · ak ) = ( a1 ak )( a1 ak−1 ) · · · ( a1 a2 )
Just look carefully to see that this works!
Example
(17645) = (15)(14)(16)(17).
Definition 9.7. σ ∈ Sn is even if it can be written as the product of an even number of transpositions.
σ may similarly be considered odd.
Theorem 9.8. The concept of evenness/oddness is well-defined: every permutation is either even or odd, and
not both.
Proof. Recall that any permutation σ ∈ Sn can be written as an n × n permutation matrix (Definition
8.6). A 2-cycle is a permutation matrix which swaps two rows: it therefore differs from the n × n
identity matrix only in that two of its rows are swapped. For example


1 0 0 0
0 0 0 1

(24) = 
 0 0 1 0  ∈ S4 .
0 1 0 0
From linear algebra we have that swapping two rows of a matrix changes the sign of its determinant.
Hence det(2-cycle) = −1. It follows that
(
1 if σ is the product of an even number of 2-cycles,
det(σ) =
−1 if σ is the product of an odd number of 2-cycles.
In particular σ cannot be both odd and even.
The concept of even permutations leads us to the definition of new collection of groups.
37
Definition 9.9. The alternating group An (n ≥ 2) is the group of even permutations in Sn .
Theorem 9.10. An has exactly half the elements of Sn : that is | An | =
n!
2.
Proof. Since n ≥ 2, we have (12) ∈ Sn . Define φ : An → {odd permutations} by φ(σ) = (12)σ. We
claim that this is a bijection
1–1 φ(σ) = φ(τ ) =⇒ (12)σ = (12)τ =⇒ στ. X
onto If ρ is an odd permutation, then (12)ρ is even and so in An . Therefore ρ = φ((12)ρ). X
Since φ is a bijection, it follows that there are exactly the same number of even and odd permutations
in Sn . Exactly half of them are therefore even.
Examples: small alternating groups
1. A2 = {e} is extremely boring!
2. A3 = {e, (13)(12), (12)(13)} = {e, (123), (132)} is simply the cyclic group of order 3.
3. A4 = {e, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)} is the
first genuinely new group in the alternating series. It has order 12 and is non-Abelian: for
example
(123)(124) = (13)(24) 6= (14)(23) = (124)(123)
We already know of one non-Abelian group of order 12: the dihedral group D6 . But we can
quickly see that A4 is non-isomorphic to D6 : all elements of A4 have order 1, 2 or 3, while D6 ,
being the symmetries of a hexagon, contains a rotation of order 6.
A concrete appearance of A4 can be seen as the group of rotations of a tetrahedron. Either label
the corners of a tetrahedron, or the faces, with the numbers 1, 2, 3, 4. Can you visualize how
each element of A4 transforms each tetrahedron?
38
10
Cosets & Lagrange’s Theorem
Take a look at any of the subgroup relations in the notes and observe that the order of every subgroup
is a divisor of the order of its parent group. This is a general result which we shall prove shortly.
Theorem 10.1 (Lagrange). Suppose that G is a finite group. The order of any subgroup H ≤ G divides the
order of G. Otherwise said
H ≤ G =⇒ | H | | G |
Remarks
• We have already seen the special case of Lagrange’s Theorem for cyclic groups (Theorem 6.6,
6.7 and Corollary 6.8).
• The Theorem is sometimes misremebered as ‘the order of an element divides the order of a
group.’ This is really only a statement about the cyclic subgroups of a group G and is not as
general as Lagrange-proper.
• Lagrange’s Theorem only makes sense for finite groups. Suppose we tried to extend it to infinite groups. Consider, for example, the fact that the rotations of an equilateral triangle form a
subgroup of the rotations of a circle. In the language above, this would yield the strange claim
that 3 | 2ℵ0 .
For such a simple result, Lagrange’s Theorem is extremely powerful. For instance, you may have
observed that there is only one group structure, up to isomorphism, of prime orders 2, 3, 5 and 7.
This is also a general result.
Corollary 10.2. There is only one group (up to isomorphism) of each prime order p, namely C p .
Proof. Let p be prime and suppose that G is a group with | G | = p.
Since p ≥ 2, we may choose some element x ∈ G \ {e}.
Now consider the cyclic subgroup generated by x, namely h x i ≤ G.
By Lagrange, we see that |h x i| divides p. Since p is prime we conclude that h x i = 1 or p.
However h x i has at least 2 elements (x and e) and so |h x i| = p.
It follows that G = h x i is cyclic.
Now we come to the ticklish business of proving Lagrange’s Theorem. The idea is essentially the
same as the way we proved Theorem 9.10. Given a subgroup H ≤ G, we partition G into subsets,
each of which has the same cardinality as H. These subsets are the cosets of H.
Definition 10.3. Let H ≤ G and choose g ∈ G. The subsets of G defined by
gH := { gh : h ∈ H }
Hg := {hg : h ∈ H }
are (respectively) the left and right cosets of H containing g.
Even though our ultimate purpose is to prove Lagrange’s Theorem (a statement about finite groups),
the concept of cosets is perfectly well-defined for any group.
39
Lemma 10.4. If G is an Abelian group then the left and right cosets of any subgroup H ≤ G are identical: i.e.
∀ g ∈ G, gH = Hg
For non-Abelian groups this is generally not true.14
It is important to note that the cosets of a subgroup H ≤ G are themselves subsets of G. Writing
out a few examples is essential to getting used to the concept!
Examples
1. Consider the subgroup of (Z12 , +12 ) generated by 4. This subgroup is cyclic of order 3:
C3 = h4i = {0, 4, 8}
In this example we will write the cosets additively, since addition is the group operation: that
is, the left coset of h4i containing x ∈ Z12 is the subset
x + h4i = { x + n : n ∈ h4i} = { x, x + 4, x + 8}
where we might have to reduce x + 4 and x + 8 modulo 12. The left cosets of h4i are as follows:
0 + h4i = {0, 4, 8}
1 + h4i = {1, 5, 9}
2 + h4i = {2, 6, 10}
3 + h4i = {3, 7, 11}
= 4 + h4i = 8 + h4i
= 5 + h4i = 9 + h4i
= 6 + h4i = 10 + h4i
= 7 + h4i = 11 + h4i
(usually written h4i, dropping the zero)
There are only four distinct cosets: notice that each has the same number of elements (3) as the
Z
subgroup h4i, and that ||h412i|| = 12
3 = 4. The cosets have partitioned Z12 into equal-sized subsets.
This is crucial for the proof of Lagrange’s Theorem.
Observe that the right cosets of h4i are the same as the left cosets. Here we have an Abelian
group and so left and right cosets are always the same (Lemma 10.4).
2. Recall the multiplication table for S3 given in the introduction. With a little work, you should
be able to compute that the left and right cosets of the subgroup H = {e, µ1 } are as follows:
Left cosets
eH = µ1 H = H = {e, µ1 }
ρ1 H = µ3 H = { ρ1 , µ3 }
ρ2 H = µ2 H = { ρ2 , µ2 }
Right cosets
He = Hµ1 = H = {e, µ1 }
Hρ1 = Hµ2 = {ρ1 , µ2 }
Hρ2 = Hµ3 = {ρ1 , µ3 }
Note that the left and right cosets of H are different this time.
3. Consider the subgroup An ≤ Sn . As we saw in Theorem 9.10,
(12) An = Bn = odd permutations in Sn
Indeed, it should not be hard to convince yourself that if σ ∈ Sn , then the cosets of An contiaining σ are
(
An if σ even
σAn = An σ =
Bn if σ odd
14 Although
see example 3 below.
40
The following Theorem is the critical step in proving Lagrange’s Theorem.
Theorem 10.5. The left cosets of H partition G.
Before reading the proof, look at each of the above examples and convince yourself that the Theorem is satisfied in each case. The strategy of the proof is to define an equivalence relation, the
equivalence classes of which are precisely the left cosets of H.
Proof. Define a relation ∼ on G by x ∼ y ⇐⇒ x −1 y ∈ H. We claim that ∼ is an equivalence relation.
Reflexivity x ∼ x since x −1 x = e ∈ H.
X
Symmetry x ∼ y =⇒ x −1 y ∈ H =⇒ ( x −1 y)−1 ∈ H, since H is a subgroup. But then
y−1 x ∈ H =⇒ y ∼ x. X
Transitivity If x ∼ y and y ∼ z then x −1 y ∈ H and y−1 z ∈ H. But H is closed, whence
x −1 z = ( x −1 y)(y−1 z) ∈ H =⇒ x ∼ z. X
The equivalence classes of ∼ therefore partition G. We claim that these are precisely the left cosets of
H. Specifically, we claim that
x ∼ y ⇐⇒ x and y lie in the same left coset of H ⇐⇒ xH = yH
However, note that x ∼ y ⇐⇒ x −1 y ∈ H ⇐⇒ y ∈ xH. It follows that yH ⊆ xH.
By symmetry, x ∼ y ⇐⇒ y−1 x ∈ H ⇐⇒ xH ⊆ yH. We conclude that
x ∼ y ⇐⇒ xH = yH
as required.
Proof of Lagrange’s Theorem. Suppose that H ≤ G and g ∈ G is a fixed element. Then the function
φ : H → gH given by φ(h) = gh is a bijection.15 It follows that every left coset of H has the same
cardinality as H. Therefore
| G | = (number of left cosets of H) · | H |
Thus | H | divides | G |.
Remarks
• The right cosets of a subgroup H also partition G although, in general, they do this in a different
way to the left cosets: observe the earlier example of the cosets of the subgroup {e, µ1 } ≤ S3 .
The proof of Lagrange’s Theorem could also be argued with right cosets.
15 Argument
exactly like that in Theorem 9.10.
41
• Each of the three conditions that ∼ be an equivalence relation corresponds to one of the properties that characterises H as being a subgroup of G: reflexivity says that H contains the identity,16
symmetry that H is closed under inverses, and transitivity that H is closed under the group
operation. It is exactly the fact that H is a subgroup that guarantees partition—if H is merely a
subset of G, we are not guaranteed a partition. For example consider the subset {0, 1} ⊂ Z3 : its
left ‘cosets’ are
0 + {0, 1} = {0, 1},
1 + {0, 1} = {1, 2},
2 + {0, 1} = {2, 1},
which are certainly not disjoint and therefore do not partition Z3 .
• It is a corollary of the proof of Lagrange’s Theorem that the number of left and right cosets of
G
H in G is identical: both are equal to the quotient ||H|| . Indeed this leads us to the following
definition.
Definition 10.6. If H ≤ G then the index of H in G, written ( G : H ), is the number of left (or right)
cosets of H in G.
Indices in Infinite Groups The concept of index makes perfect sense for infinite groups, although
it is much less intuitive. One defines ( G : H ) to be the cardinality of the set of left (or right) cosets of H
in G. It is theorem that these cardinalities are always identical. Moreover, the argument in the proof
of Lagrange’s Theorem that all cosets of a subgroup have the same cardinality is perfectly valid, even
if G is infinite. The difficulty is that we cannot divide infinite cardinals. For example it can be shown
that
(On (R) : SOn (R)) = 2
Intuitively this says that half of the length/angle-preserving linear transformations of Rn are rotations and half reflections. This may seem strange since both groups have the same cardinality:
ℵ
|On (R)| = |SOn (R)| = 2ℵ0 . It would be absurd to write 22ℵ00 = 2.
A similar challenge to our intuition follows from
((Q, +) : (Z, +)) = ℵ0
where both Q and Z have the same cardinality ℵ0 . Again, ℵℵ00 = ℵ0 is an absurdity.
In this course, you are only expected to be completely comfortable with indices in finite groups!
Theorem 10.7. If K ≤ H ≤ G is a sequence of subgroups such that ( G : H ) and ( H : K ) are both finite, then
( G : K ) = ( G : H )( H : K )
By the above Remarks, if all three groups were finite, then the argument is trivial:
(G : K) =
16 Strictly
|G|
|G| | H |
=
·
= ( G : H )( H : K )
|K |
| H | |K |
this comes for free given the other two, recall Theorem 5.2.
42
In the proof, we include the possibility that G, H and K are infinite groups; it is just the indices that
must be finite.
The proof is, in essence, very similar to the proof that the cardinality of a Cartesian product of sets
is the product of the cardinalities of the sets: | A × B| = | A| | B|. The challenge is twofold: finding/labelling the objects (cosets) we wish to count, and justifying that they really are all the cosets of
interest.
Proof. If ( G : H ) and ( H : K ) are finite, then they are necessarily natural numbers. Thus let p = ( G :
H ) and q = ( H : K ) where p, q ∈ N. We may therefore enumerate the left cosets of H in G: ∃ distinct
gi ∈ G such that these are
g1 H, g2 H, . . . , g p H.
We may also enumerate the left cosets of K in H: ∃ distinct h j ∈ H such that these cosets are
h1 K, h2 K, . . . , hq K.
We claim that the left cosets of K in G are given by ( gi h j )K where i = 1, . . . , p and j = 1, . . . , q.
Essentially
g1 h 1 K
g2 h 1 K
..
.
g1 h 2 K · · ·
g2 h 2 K · · ·
..
..
.
.
g p h1 K g p h2 K · · ·
g1 h q K
g2 h q K
..
.
g p hq K
If our claim is correct, then there are clearly pq left cosets of K in G.
To prove the claim, first note that each set gi h j K is a left coset of K in G. If we can show that the above
list of cosets partitions G, then we know that these are all of the required left cosets. We therefore need
to see that every element of G lies in exactly one of these cosets.
1. Let x ∈ G be given. Since the left cosets of H partition G we see that every x lies in exactly
one left coset x ∈ gi H. Therefore gi−1 x ∈ H. However every element of H lies in exactly one
left coset h j K. It follows that ∃i, j such that x ∈ ( gi h j )K. Every x ∈ G is therefore in one of the
claimed cosets of K.
2. Now suppose y ∈ gi h j K ∩ gα h β K for some i, j, α, β. Then y ∈ gi H ∩ gα H =⇒ α = i, since the
left cosets of H partition G. But then gi−1 x ∈ h j K ∩ h β K =⇒ β = j since the left cosets of K in
H partition H. It follows that the cosets ( gi h j )K are disjoint.
By 1 and 2, the left cosets ( gi h j )K partition G. It follows that these are precisely the left cosets of K in
G, and the result follows.
Example In case the proof makes you nervous, here is a concrete example where we describe the
cosets in the same language.
Consider G = Z20 with its subgroup H = h2i. There are two (left)17 cosets of H in G, namely
H = {0, 2, 4, . . . , 18}
and
1 + H = {1, 3, 5, . . . , 19}
17 The example is Abelian so left and right cosets are identical: recall Lemma 10.4. Remember that the notation for this
example is additive!
43
In the language of the proof, p = ( G : H ) = 2 with g1 = 0 and g2 = 1.
Meanwhile, H has a subgroup K = h10i = {0, 10}, which has five cosets in H:
K = {0, 10},
2 + K = {2, 12},
4 + K = {4, 14},
6 + K = {6, 16},
8 + K = {8, 18}
In this case we have q = ( H : K ) = 5 with h1 = 0, h2 = 2, h3 = 4, h4 = 6, h5 = 8.
There are ten cosets of K in G:
K = {0, 10},
1 + K = {1, 11},
2 + K = {2, 12},
...,
9 + K = {9, 19}
Notice that these can be written
K = ( g1 + h1 ) + K,
1 + K = ( g2 + h1 ) + K,
...,
2 + K = ( g1 + h2 ) + K,
8 + K = ( g1 + h4 ) + K,
...
9 + K = ( g2 + h 4 ) + K
The cosets are precisely as constructed in the proof.
11
Factor Groups
(At this point we diverge markedly from the order of things in Fraleigh: this section corresponds to
part of chapter 14 in the textbook)
The idea of factor groups is very simple, if abstract. Given a subgroup H ≤ G we may consider
the set of left cosets18
G = { gH : g ∈ G }
H
The question for this section is whether the set of left cosets has any interesting structure. In the context
of group theory, this means can we view G H as a group in a natural way?
A natural structure? To make the set of left cosets into a group we need a binary operation. This
might be done very easily. Suppose, for instance, that ( G : H ) = 4. We could simply place the left
cosets in any order we like g1 H, g2 H, g3 H, g4 H and define a binary operation ∗ on G H by
gi H ∗ g j H := gi+4 j H.
The function φ : G H , ∗ → (Z4 , +4 ) is then a group isomorphism. The problem is one of choice.
There is nothing to stop us ordering the cosets of H differently. Or indeed, we could define an iso
morphism of G H with the Klein 4-group instead. Any group structure imposed on G H defined in
such an arbitrary way will tell us very little about G itself. Such a structure is essentially useless.
18 Recall
(Theorem 10.5) that the left cosets of H partition G, and that this is done by means of an equivalence relation ∼:
x ∼ y ⇐⇒ x −1 y ∈ H ⇐⇒ xH = yH
The equivalence classes of ∼ are precisely the left cosets of H in G. In previous classes you would have written the set of
equivalence classes as G ∼ . In this context, when the equivalence relation is defined by the subgroup H, it is simpler to
ignore ∼ entirely and refer instead to G H .
44
A natural operation, by contrast, means that the binary operation on G H comes directly from the
group operation of G, independent of any choices on our part. Any such natural group structure will
necessarily encode information about G.
To see how this works, let us recall the first two example of cosets on page 40.
Turning the set of left cosets into a group
cosets
H,
1 + H,
2 + H,
Consider H ≤ Z12 where H = h4i = {0, 4, 8}, with (left)
3+H
The set of left cosets is then
n
o
Z12 = { H, 1 + H, 2 + H, 3 + H } = {0, 4, 8}, {1, 5, 9}, {2, 6, 10}, {3, 7, 11}
H
There is a nice pattern here: x and y are members of the same coset if and only if x ≡ y (mod 4).
Moreover, there are four cosets. Can we view Z12 H as merely the cyclic group Z4 in disguise? The
answer, of course, is yes.
To do this, we need to define a binary operation on Z12 H : we need to see how to combine cosets
to create new ones. The obvious thing to do is to use the addition in Z12 that we already have: thus
define
( a + H ) ⊕ (b + H ) := ( a + b) + H
The binary operation ⊕ on the set Z12 H comes naturally from the existing addition on Z12 : we
haven’t created anything new, we’re just using what we’ve already been given.
There is a potential problem however: the freedom of choice built into the definition. Computing
( a + H ) ⊕ (b + H ) requires three steps:
Choose representatives Make a choice of elements a and b in the respective cosets.
Add in the original group Compute a + b ∈ Z12 .
Take the coset Return the left coset ( a + b) + H.
If ⊕ is to make any sense at all, then the result ( a + b) + H must be independent of the choices we
made in the first step.19 In this case there is no problem, as you can tediously check for yourself. For
example, to compute
(1 + H ) ⊕ (2 + H )
19 Strictly
speaking we require that ⊕ : Z12 H × Z12 H → Z12 H be a well-defined function.
45
there are nine possibilities:
1+2 = 3
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
1+6 = 7
1 + 10 = 11
5+2 = 7
5 + 6 = 11
5 + 10 = 15
9 + 2 = 11
9 + 6 = 15
9 + 10 = 19
(1 + H ) ⊕ (2 + H ) = 3 + H
(1 + H ) ⊕ (2 + H ) = 7 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 11 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 7 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 11 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 15 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 11 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 15 + H = 3 + H
(1 + H ) ⊕ (2 + H ) = 19 + H = 3 + H
Thankfully all nine results are the same! The other possibilities for adding cosets can be similarly
checked (we’ll do this in general in a moment). It is not hard to produce a table for the binary
operation ⊕ on the set Z12 H :
⊕
H
1+H
2+H
3+H
H
H
1+H
2+H
3+H
1+H
1+H
2+H
3+H
H
2+H
2+H
3+H
H
1+H
3+H
3+H
H
1+H
2+H
This table should look very familiar: if you delete all the H expressions (so that H = 0 + H becomes 0), we recover precisely the Cayley table for the cyclic group Z4 . Indeed we have proved the
following:
Proposition. The set of left cosets Z12 H = { H, 1 + H, 2 + H, 3 + H } forms a group under the operation
⊕, which is moreover isomorphic to Z4 . We will shortly refer to this as a factor group.
It would be very nice if this sort of behavior always happened. Unfortunately it doesn’t.
When cosets fail to form a group
Recall that its left cosets are
eH = µ1 H = H = {e, µ1 },
Let us repeat the process with the subgroup H = {e, µ1 } ≤ D3 .
ρ1 H = µ3 H = { ρ1 , µ3 },
ρ2 H = µ2 H = { ρ2 , µ2 }
In the same way as above, we define the ‘natural’ operation on the set D3 H of left cosets:20
aH ⊗ bH := ( ab) H
This time, there is a problem. Suppose we wanted to compute ρ1 H ⊗ ρ1 H. There are four choices:21
ρ1 H ⊗ ρ1 H = ρ21 H = ρ2 H
ρ1 H ⊗ µ3 H = ρ1 µ3 H = µ2 H = ρ2 H
µ3 H ⊗ ρ1 H = µ3 ρ1 H = µ1 H = H
µ3 H ⊗ µ3 H = µ23 H = eH = H
20 Again,
21 Write
the ‘operation’ on the set of left cosets comes directly from the group operation on D3 .
these as cycles if you’re having trouble computing: e.g. ρ1 = (123), µ1 = (23), etc.
46
Exercising the freedom of choice in the definition of ⊗ leads to different outcomes: it follows that ⊗ is
not well-defined. There is nothing stopping us from defining a group structure on D3 H , but any such
definition must involve a choice on our part: we would be imposing structure rather that observing
structure that occured naturally.
Well-definition of the natural group structure The above discussion can be generalized. Clearly
some subgroups H ≤ G behave nicely with respect to the natural structure on the set G H of left
cosets. How can we tell which subgroups?
Let H be a subgroup of G and define the natural operation on G H :
aH · bH := ( ab) H
This is well-defined iff
∀ a, b ∈ G, ∀ x ∈ aH, y ∈ bH,
we have
aH · bH = xH · yH
Let us trace through what this means for the subgroup H.
x ∈ aH ⇐⇒ ∃h ∈ H such that x = ah
The operation on G H is therefore well-defined if and only if
∀ a, b ∈ G, ∀h1 , h2 ∈ H, ( ah1 bh2 ) H = ( ab) H
⇐⇒ ∀ a, b ∈ G, ∀h ∈ H,
( ahb) H = ( ab) H
⇐⇒ ∀ a, b ∈ G, ∀h ∈ H,
⇐⇒ ∀b ∈ G,
⇐⇒ ∀b ∈ G,
⇐⇒ ∀b ∈ G,
∀h ∈ H,
∀h ∈ H,
∀h ∈ H,
( ab)−1 ( ahb) ∈ H
(since h2 H = H for all h2 ∈ H)
b−1 hb ∈ H
hb ∈ bH
Hb = bH
It follows that G H is a group in a natural way if and only if the left and right cosets of H in G are
identical. Our argument above leads to a definition and constitutes the proof of the next theorem.
Definition 11.1. Let H be a subgroup of G. We say that H is a normal subgroup of G if its left and right
cosets are identical. We write H / G.
Theorem 11.2. The following are equivalent:
1. H / G is a normal subgroup.
2. ∀ g ∈ G, gH = Hg.
3. ∀ g ∈ G, h ∈ H, ghg−1 ∈ H.
4. ∀ g ∈ G, gHg−1 = H.
Given that the whole reason for defining normal subgroups was our desire for G H to form a
group in a natural way, we are led to the following:
47
Theorem 11.3. Suppose that H / G is a normal subgroup. Then the set of (left) cosets G H forms a group
under the natural operation
aH · bH := ( ab) H
Proof. We check the group axioms.
Closure By the above discussion, H / G ⇐⇒ the natural operation is well defined. Certainly
aH · bH = ( ab) H is a coset, whence G H , · is closed. X
Associativity aH · (bH · cH ) = aH · (bc) H = a(bc) H. Similarly ( aH · bH ) · cH = ( ab)cH. By the
associativity of G these are identical. X
Identity eH · aH = (ea) H = aH therefore eH = H is the identity. X
Inverse a−1 H · aH = ( a−1 a) H = eH = H, therefore ( aH )−1 = a−1 H. X
Warning! Because the group structure on a factor group G H comes naturally from the group structure on G, we typically use the same notation for the operation. Thusif ( G, +)
is a group, we also use +
G
G
for the operation on H . Similarly ( G, ·) will have factor groups
H , · written multiplicatively.
Make sure you know to which group an operation refers! The symbols ⊕, ⊗ in the above examples
were used only to help you keep the operations separate.
The group Zm
There are many nice examples of factor groups. This next is important enough that it merits a full
discussion.
For each integer m, its integer multiples mZ = hmi form a subgroup of Z. Since Z is Abelian,
mZ is a normal subgroup. We may write the cosets of mZ as
Z
mZ = {0 + hmi , 1 + hmi , . . . , (m − 1) + hmi}.
Addition in the factor group Z mZ is then
i + h m i + j + h m i = (i + j ) + h m i .
This looks suspiciously like addition modulo m. Indeed we have the following:
Theorem 11.4. µ : Z mZ → Zm : x + hmi 7→ x (mod m) is an isomorphism.
Proof.
Well-definition and Injectivity Note that22
x + hmi = y + hmi ⇐⇒ x − y ∈ hmi ⇐⇒ x − y ≡ 0
22 A
⇐⇒ µ( x + hmi) = µ(y + hmi)
(mod m) ⇐⇒ x = y (in Zm )
function f is well-defined if ∀ x, y ∈ dom( f ), x = y =⇒ f ( x ) = f (y). This is precisely the converse of injectivity.
48
Surjectivity Given any x ∈ Zm , we have x = µ( x + hmi).
Homomorphism For all x + hmi , y + hmi ∈ Z mZ , we have
µ ( x + hmi) + (y + hmi) = µ ( x + y) + hmi = x + y (mod m)
= µ x + hmi + µ y + hmi (mod m)
Given this argument, many algebraists simply take the factor group Z mZ as the definition of the
group (Zm , +m ). This obviates any need to use congruence arithmetic.
A very similar result applies to finite cyclic groups: recall (Theorem 6.7) that hsi ≤ Zn has order
d = gcdn(s,n) and that there is precisely one subgroup of Zn of order d for each divisor d of n.
Theorem 11.5. µ : Zn hsi → Zgcd(s,n) : x + hsi 7→ x (mod gcd(s, n)) is an isomorphism. Alternatively,
if d | n, then
Cn ∼
Cd = C nd
Example
h5i ≤ Z20 has cosets
Z20
h5i = {0 + h5i , 1 + h5i , 2 + h5i , 3 + h5i , 4 + h5i},
which is isomorphic to Z5 under the isomorphism
µ : Z20 h5i → Z5 : x + h5i 7→ x
(mod 5)
More Examples
1. We can do something similar with cyclic subgroups of the real numbers under addition. In
particular, h2π i = {2πn : n ∈ Z} is such a subgroup. Given that (R, +) is Abelian, it follows
that h2π i is normal and that R h2π i is a factor group. How can we visualize this group?
If you think carefully, there is precisely one coset for each value in the half-open interval [0, 2π ).
The factor group R h2π i can be visualized as the unit circle in the complex plane under multiplication. Explicitly, we define
µ : R h2π i → S1 : x + h2π i 7→ eix
and claim that this is an isomorphism.
Well-definition and Injectivity Note that
x + h2π i = y + h2π i ⇐⇒ x − y ∈ h2π i ⇐⇒ ∃n ∈ Z : x − y = 2πn
⇐⇒ ei(x−y) = 1 ⇐⇒ eix = eiy
⇐⇒ µ( x + h2π i) = µ(y + h2π i)
49
Surjectivity All point in S1 ⊆ C have the form eix = µ( x + h2π i) for some angle x.
Homomorphism For all x + h2π i , y + h2π i ∈ R h2π i , we have
µ ( x + h2π i) + (y + h2π i) = µ ( x + y) + h2π i = ei(x+y) = eix · eiy
= µ x + h2π i · µ y + h2π i
We conclude that R h2π i , + ∼
= (S1 , ·).
2. Recall the description of the alternating group A4 . It can be seen that the
V = {e, (12)(34), (13)(24), (14)(23)}
is a subgroup of A4 which is moreover isomorphic to the Klein 4-group. It can be see that V / A4
is a normal subgroup. In this case, the factor group A4 V is C3 .
3. A similar analysis may be performed to see that S4 V ∼
= S3 . See the homework.
12
Homomorphisms and the First Isomorphism Theorem
In each of our examples of factor groups, we not only computed the factor group but identified it as
isomorphic to an already well-known group. Each of these examples is in fact a special case of a very
important theorem: the first isomorphism theorem. This theorem provides a universal way of defining
and identifying factor groups G H . Moreover, it has versions applied to all manner of algebraic
structures, the most famous being the rank–nullity theorem in linear algebra. In order to discuss this
theorem, we need to consider two subgroups related to any group homomorphism φ : G → L.
Definition 12.1. Let φ : G → L be a homomorphism of multiplicative groups. The kernel and image
of φ are the sets
ker φ = { g ∈ G : φ( g) = e L }
Im φ = {φ( g) : g ∈ G }
Note that ker φ ⊆ G while Im φ ⊆ L. This is the usual notion for the image of a function: Im φ =
range φ.
Theorem 12.2. Suppose that φ : G → L is a homomorphism. Then
1. φ(eG ) = e L .
2. ∀ g ∈ G, (φ( g))−1 = φ( g−1 ).
3. ker φ / G.
4. Im φ ≤ L.
Proof.
1. Suppose that g ∈ G. Then
φ( g) = φ( geG ) = φ( g)φ(eG ) =⇒ e L = φ(eG )
by the left cancellation law.
50
2. Suppose that g ∈ G. Then
e L = φ(eG ) = φ( gg−1 ) = φ( g)φ( g−1 ) =⇒ (φ( g))−1 = φ( g−1 )
3. First suppose that k1 , k2 ∈ ker φ. Then
φ(k1 k2 ) = φ(k1 )φ(k2 ) = e L =⇒ k1 k2 ∈ ker φ
and
1
−1
1
φ(k−
= e L =⇒ k−
1 ) = ( φ ( k 1 ))
1 ∈ ker φ
It follows that ker φ is a subgroup of G. To see that ker φ is normal, recall Theorem 11.2. Let
g ∈ G and k ∈ ker φ. Then
φ( gkg−1 ) = φ( g)φ(k )φ( g)−1 = e L =⇒ gkg−1 ∈ ker φ
It follows that ker φ / G.
4. Let φ( g1 ), φ( g2 ) ∈ Im φ. Then
φ( g1 )φ( g2 ) = φ( g1 g2 ) ∈ Im φ
and
(φ( g1 ))−1 = φ( g1−1 ) ∈ Im φ
Thus Im φ is a subgroup of L.
Examples
1. Recall that the trace function tr : Mn (R) → R is a homomorphism of additive groups. These
are Abelian groups and so the kernel of tr is automatically normal without needing the above
Theorem. The additive group of trace-free matrices23 ) is a normal subgroup of ( Mn (R), +):
ker tr = { A ∈ Mn (R) : tr A = 0} / Mn (R).
2. φ : Z36 → Z20 defined by φ(n) = 4n (mod 24). The kernel of φ is the subgroup
ker φ = {n : 4n ≡ 0
(mod 24)} = {0, 6, 12, 18, 24, 30} / Z36
is the cyclic group C6 .
(
1
if σ even,
is a homomorphism of
−1 if σ odd,
groups. Here {1, −1} is a group under multiplication. Since the identity in the target group is
1, we have ker sgn = An , the alternating group of even permutations in Sn . Indeed An / Sn .
3. The map sgn : Sn → {1, −1} given by sgn(σ) =
23 This
kernel is often written sln (R) (for the special-linear algebra), and is very much related to the special linear group
SLn (R): the expression det e A = etr A shows that matrix exponentiation is a map exp : sln (R) → SLn (R). This is the
beginning of the theory of Lie groups and their Lie algebras.
51
4. det : GLn (R) → R× is a homomorphism and so ker det = SLn (R) is a normal subgroup
SLn (R) / GLn (R). Compare this with example 1.
All these examples should suggest an idea to you. Not only are all kernels of homomorphisms
normal subgroups, the converse is also true: any normal subgroup is in fact the kernel of some homomorphism of groups. This (loosely) is the 1st isomorphism Theorem which we will come to shortly.
The homomorphism approach makes calculations with factor groups easier! As the next lemma
shows, there is a very easy correspondence between the cosets of the kernel of a homomorphism,
and the elements of the image.
Lemma 12.3. Suppose that φ : G → L is a homomorphism. Then
g1 ker φ = g2 ker φ ⇐⇒ φ( g1 ) = φ( g2 )
Proof. For all g1 , g2 ∈ G, we have
g1 ker φ = g2 ker φ ⇐⇒ g2−1 g1 ∈ ker φ
⇐⇒ φ( g2−1 g1 ) = e L
⇐⇒ φ( g2 )−1 φ( g1 ) = e L
⇐⇒ φ( g1 ) = φ( g2 )
The lemma says that there is a bijective correspondence between the factor group G ker φ and
the image Im φ. In the next theorem, we put this to use to help us determine what can possibly be a
homomorphism.
Theorem 12.4. Let φ : G → L be a homomorphism.
1. If G is a finite group then
Im φ is a finite subgroup of L and its order divides that of G. More succinctly:
| G | < ∞ =⇒ |Im φ| | G |.
2. Similarly, | L| < ∞ =⇒ |Im φ| | L|.
Note that we are only assuming one of the groups to be finite in each case.
Proof.
1. If G is a finite group, then Im φ is a finite subgroup of L. Lemma 12.3 tells us that
φ( g1 ) = φ( g2 ) ⇐⇒ g1 ker φ = g2 ker φ
whence there are precisely as many elements of Im φ as there are distinct cosets of ker φ in G.
But this is the definition of the index:
|Im φ| = ( G : ker φ)
Given that G is finite, the discussion following Lagrange’s Theorem quickly says that
| G | = |Im φ| · |ker φ| =⇒ |Im φ| | G |
2. This is immediate from Lagrange’s Theorem: Im φ is a subgroup of a finite group L and so the
order of Im φ divides the order of L.
52
Examples
1. How many homomorphisms are there φ : G → L when G and L have relatively prime orders?
If φ were such a homomorphism, then the Theorem says that |Im φ| divides both | G | and | L|.
But the only such positive integer is 1. Since the image of any homomorphism always contains
the identity (φ(eG ) = e L ), it follows that there is only one homomorphism! φ must be the
function φ : g 7→ e L , ∀ g ∈ G.
2. How many homomorphisms are there φ : Z4 → S3 ? Again the Theorem tells us that |Im φ|
divides 4 = |Z4 | and 6 = |S3 |: thus |Im φ| is either 1 or 2. If |Im φ| = 2 then Im φ is a subgroup
of order 2 of S3 . There are exactly 3 of these: {e, µi } for each i = 1, 2, 3. Since a homomorphism
must map the identity to the identity we therefore have the homomorphisms
 
0
1

φ1 : 
2 7 →
3

e
 µ1 
 ,
e
µ1

 
0
1

φ2 : 
2 7 →
3

e
 µ2 
 ,
e
µ2

 
0
1

φ3 : 
2 7 →
3

e
 µ3 
 .
e
µ3

Finally if Im φ has one element then φ is the trivial homomorphism φ(z) = e, ∀z ∈ Z4 . There
are therefore 4 distinct homomorphisms.
Proposition 12.5. There are exactly d = gcd(m, n) distinct homomorphisms φ : Zm → Zn , defined by
n
φ( x ) = k x
d
(mod n) where k = 0, . . . , d − 1.
There are several ways of proving this: one is in the homework. We use the above discussion on
the cardinality of the image.
Proof. Suppose that φ : Zm → Zn is a homomorphism. Then |Im φ| d. However Im φ is a subgroup
of Zn which is necessarily cyclic. Recalling the structure of cyclic groups24 we see that must be a
subgroup of the unique cyclic subgroup of order d in Zn . This is generated by nd . The only possible
choices for our homomorphisms are therefore25
n
φk ( x ) = k x
d
(mod n)
for each k = 0, 1, 2, . . . , d − 1. It remains only to check that these are well-defined functions. For this,
note that for any j ∈ Z we have,
n
m
n
n
φk ( x + jm) = k ( x + jm) ≡ k x + knj ≡ k x
d
d
d
d
= φk ( x )
(mod n)
(since
m
d
∈ Z)
24 Recall that a cyclic group Z has exactly one subgroup, which is itself cyclic, of each order d which divides n. Therefore
n
if Im φ is a subgroup of Zn , then it is cyclic. Since its order divides d, it must also be a subgroup of the unique Cd ≤ Zn .
25 We may choose φ (1) for each k. If these are homomorphisms, then φ (2) = φ (1) + φ (1), etc.
k
k
k
k
53
Example Suppose that φ : Z12 → Z16 is a homomorphism. Since gcd(12, 16) = 4, the image of φ
must be a subgroup of C4 = h4i ≤ Z16 . There are four choices:
φ0 ( x ) = 0,
φ1 ( x ) = 4x,
φ2 ( x ) = 8x,
φ3 ( x ) = 12x
(mod 16).
Reversing the argument, we see that there are also four distinct homomorphisms ψ : Z16 → Z12 ,
namely
ψ0 ( x ) = 0,
ψ1 ( x ) = 3x,
ψ2 ( x ) = 6x,
ψ3 ( x ) = 9x
(mod 12).
We have seen that all kernels of group homomorphisms are normal subgroups. In fact all normal
subgroups are the kernel of some homomorphism. We state this as a 2-part Theorem.
Theorem 12.6 (1st Isomorphism Theorem).
1. Let H / G. Then γ : G → G H defined by γ( g) = gH
is a homomorphism with ker γ = H.
2. If φ : G → L is a homomorphism with kernel H, then µ : G H → Im φ defined by µ( gH ) = φ( g) is
an isomorphism.
The Theorem can be summarised by the following formula,
G
∼
ker φ = Im φ.
In advanced algebra, the notion of a commutative diagram is useful for
summarizing theorems. To say that a diagram commutes means that
if there are multiple paths from one side of the diagram to the other,
they both give the same result. In this case one can travel from G to
Im φ in two ways. For the 1st isomorphism theorem, this means that
φ = µ ◦ γ.
φ
G
γ
!
/ Im φ
;
µ
G
ker φ
Definition 12.7. Given a normal subgroup H / G, the homomorphism γ : G → G H defined above
is the canonical homomorphism.
Proof of Theorem. We need to check that the two functions γ and µ defined in the Theorem have the
properties we claim.
1. Let H / G and define γ : G → G H by γ( g) = gH. This is certainly a well-defined function, so
we need only check that it is a homomorphism with ker γ = H. However
γ ( g1 ) γ ( g2 ) = g1 H · g2 H = ( g1 g2 ) H = γ ( g1 g2 )
by the definition of multiplication in a factor group. Thus γ is indeed a homomorphism. Moreover, the identity in the factor group is H, whence
ker γ = { g ∈ G : γ( g) = H }
Thus g ∈ ker γ ⇐⇒ gH = H ⇐⇒ g ∈ H, and so the kernel of γ is H, as claimed.
2. Suppose φ : G → L is a homomorphism with kernel H. Then H is a normal subgroup of G and
so the factor group G H is well-defined. We need to check that µ : G H → Im φ defined by
µ( gH ) = φ( g) is an isomorphism.
54
Well-definition and Bijectivity These are immediate from Lemma 12.3.
Homomorphism For all g1 H, g2 H ∈ G H , we have
µ ( g1 H · g2 H ) = µ ( g1 g2 H ) = φ ( g1 g2 )
(since φ is a homomorphism)
= φ ( g1 ) φ ( g2 )
= µ ( g1 H ) µ ( g2 H )
µ is a well-defined, bijective homomorphism and is therefore an isomorphism µ : G H ∼
= Im φ.
Here are several examples where we can calculate all the pieces in the Theorem.
Examples
1. Let φ : Z10 → Z20 be the homomorphism with φ(1) = 4. Then, by the homomorphism property, φ( x ) = 4x (mod 20). In particular
ker φ = { x ∈ Z10 : 4x ≡ 0
(mod 20)} = {0, 5} and,
Im φ = {4x (mod 20) : x ∈ Z10 } = {0, 4, 8, 12, 16}.
Observe that ker φ = h5i ≤ Z10 is isomorphic to C2 and Im φ = h4i ≤ Z20 is isomorphic to C5 .
The factor group is
n
o
Z10
=
{
0,
5
}
,
{
1,
6
}
,
{
2,
7
}
,
{
3,
8
}
,
{
4,
9
}
.
ker φ
The function

 
{0, 5}
0
{1, 6}
4


 

 
µ : Z10 ker φ → Im φ : 
{2, 7} 7→  8 
{3, 8}
12
{4, 9}
16

i.e.
µ( x + ker φ) = 4x
(mod 20)
is the isomorphism from the Theorem.
2. Rather than starting with a homomorphism, suppose we simply wanted to identify the factor
group Z10 H with a well-known group (here H = {0, 5} is our normal subgroup from before).
We might search for a homomorphism ψ : Z10 → L whose kernel is H. Since Z10 H clearly has
5 elements (5 cosets of H in Z10 ), the obvious thing to do is search for a homomorphism
φ : Z10 → Z5
with
ker φ = H
It should be clear that φ( x ) = x (mod 5) is such a homomorphism! Since Im φ = Z5 , the first
isomorphism theorem says that Z10 H ∼
= Z5 . The isomorphism µ in this case is simply
µ( x + H ) = x
(mod 5)
55
The point of these two examples is that the Theorem can be used in two ways: both of which help
us find alternative presentations of factor groups.
• Start with a homomorphism φ : G → L and build an isomorphism µ : G ker φ ∼
= Im φ.
• Start with a normal subgroup H / G and construct a homomorphism φ with H = ker φ. Then
G ∼
H = Im φ.
13
Conjugacy, Automorphisms and Centers of Groups
In this section we describe another equivalence relation on a group, that of conjugacy. For Conjugacy
is the group theory abstraction of the idea of similarity from linear algebra: two square matrices A
and B are said to be similar if there exists some invertible matrix S for which B = SAS−1 . You should
recall that such matrices have the same eigenvalues and, essentially, ‘do the same thing’ sith respect
to different bases.
Definition 13.1. Let G be a group and g1 , g2 ∈ G. We say that g1 is conjugate to g2 if
∃h ∈ G such that g2 = hg1 h−1 .
Note that conjugation is intimately related to the concept of normal subgroups. By Theorem 11.2
a subgroup H ≤ G is normal iff
∀ g ∈ G, h ∈ H,
ghg−1 ∈ H
A normal subgroup can therefore be defined as a subgroup which is closed under conjugation.
Theorem 13.2. Conjugacy is an equivalence relation.
The proof is in the homework.
Definition 13.3. The equivalence classes of a group G under conjugacy are termed the conjugacy
classes of G.
Examples
1. If G is Abelian then every conjugacy class contains only one element. This is since g1 and g2 are
conjugate iff
∃h ∈ G such that g2 = hg1 h−1 = g1 hh−1 = g1
2. The smallest non-Abelian group is D3 ∼
= S3 . Its conjugacy classes are
{ e },
{ ρ1 , ρ2 },
{ µ1 , µ2 , µ3 }.
This can be computed directly, but it follows immediately from. . .
Theorem 13.4. The conjugacy classes of Sn are the cycle types.
56
The concept of cycle type is straightforward: for example (123)(45) has the same cycle type as
(156)(23), but not the same as (12)(34). Just write an element of Sn as a product of disjoint cycles
and its cycle type is clear. The proof involves some messy notation, so you may want to avoid on a
first reading.
Proof. Suppose that ( a1 · · · ak ) ∈ Sn is a k-cycle and ρ ∈ Sn is any element. It is easy to see that
ρ( a1 · · · ak )ρ−1 = (ρ( a1 ) · · · ρ( ak ))
is also a k-cycle. Now let τ ∈ Sn be written as the product of disjoint cycles τ = τ1 · · · τl . Then
ρτρ−1 = (ρτ1 ρ−1 )(ρτ2 ρ−1 ) · · · (ρτl ρ−1 ),
which has the same cycle type as τ.
Conversely let σ = σ1 · · · σl and τ = τ1 · · · τl be elements of Sn with the same cycle type; otherwise
said, σi and τi are k-cycles for the same k. Define a permutation π by writing σ and τ one on top of
the other in standard notation:
σ1 σ2 · · · σl s
π=
,
τ1 τ2 · · · τl t
where we write each of the elements of the cycle σi in a row and s and t are the remaining elements
of the set {1, 2, . . . , n} missing from each row written in any order. We claim that τ = πσπ −1 . If σi,j
and τi,j refer to the jth element of the orbits σi and τi respectively, then
πσπ −1 (τi,j ) = πσ (σi,j ) = πσi,j+1 = τi,j+1 = τ (τi,j ).
Example According to the Theorem, the permutations (145)(627) and (165)(234) in S7 are conjugate by the permutation
1 4 5 6 2 7 3
1 2 3 4 5 6 7
π=
=
= (23746)
1 6 5 2 3 4 7
1 3 7 6 5 2 4
Indeed
π (145)(627)π −1 = (23746)(145)(627)(26473) = (165)(234).
There are other possible choice of π, for example by writing the orbits in different orders.
Example At the end of Section 11 we claimed that V = {e, (12)(34), (13)(24), (14)(23)} is a normal
subgroup of A4 . This is clear since V contains all the elements in A4 of the cycle type ( ab)(cd) and
conjugation preserves cycle type. Thus A4 V is a factor group of order 12
4 = 3 and must therefore be
26
isomorphic to C3 . Indeed the cosets of V may be written
A4 = {V, (123)V, (132)V }.
V
An example isomorphism µ : A4 V → C3 from the first isomorphism theorem is then

 

V
e
µ : (123)V  = (123)
(132)V
(132)
where we view C3 in a natural way as a subgroup of S4 .
26 In
fact (123)V = (134)V = (243)V = (142)V and (132)V = (143)V = (234)V = (124)V.
57
Automorphisms
Conjugation by a fixed element of a group is a special case of a general structure-preserving transformation.
Definition 13.5. An automorphism of a group G is an isomorphism of G with itself. The set of such is
labelled Aut G. The inner automorphisms of G are the conjugations
Inn G = {ch : G → G where ch ( g) = hgh−1 }.
Theorem 13.6. The set Aut G and Inn G form groups under composition. Moreover Inn G is a subgroup of
Aut G.
Proof. We sketch the argument for Aut G first.
Closure It is straightforward to check that composition of two isomorphisms is an isomorphism.
Associativity Aut G consists of functions under composition: this is associative by Theorem 2.5.
Identity id : g 7→ g is clearly an automorphism and moreover ∀ψ ∈ Aut( G ),
ψ ◦ id = ψ.
Inverse If φ ∈ Aut G it is also staightforward to check that the inverse function φ−1 : G → G is
automorphism.
For Inn G, note that each ch is a homomorphism:
ch ( g1 g2 ) = h( g1 g2 )h = (hg1 h−1 )(hg2 h)−1 = ch ( g1 )ch ( g2 ).
1
It also has inverse c−
h = c h−1 since
ch−1 (ch ( g)) = h−1 (hgh−1 )h = g =⇒ ch−1 ◦ ch = id .
Thus each ch is an automorphism of G and moreover Inn G is closed under composition and inverses.
Inn G is therefore a subgroup of Aut G.
In fact Inn G is a normal subgroup of Aut G.
Theorem 13.7. Inn G / Aut G.
Proof. Let τ ∈ Aut G and ch ∈ Inn G. By Theorem 11.2 it is enough to see that τch τ −1 ∈ Inn G. Since
τch τ −1 is a function, we compute what it does to a general element g ∈ G.
(τch τ −1 )( g) = τ ch τ −1 ( g) = τ hτ −1 ( g)h−1
= τ ( h ) τ τ −1 ( g ) τ ( h −1 )
(since τ is a homomorphism)
= τ (h) gτ (h)−1
= c τ ( h ) ( g ).
(again since τ is an homomorphism)
Therefore τch τ −1 = cτ (h) , which is an element of Inn G.
Since conjugation by an element g ∈ G is an automorphism of G, conjugation must map subgroups of G to isomorphic subgroups. We also call such subgroups conjugate.
58
Example Let H = {e, µ1 } ≤ D3 . Then
cρ1 H = ρ1 Hρ1−1 = {e, ρ1 µ1 ρ2 } = {e, µ2 }.
If we compute all six possible conjugations of H, we obtain
{e, µ1 } = H = ce H = cµ1 H,
{e, µ2 } = cρ1 H = cµ3 H
{e, µ3 } = cρ2 H = cµ2 H
This comprises all of the two element subgroups of D3 .
Centers
We say that an element g in a group G commutes with another element a ∈ G if the order of multiplication is irrelevant: i.e. if ga = ag. It is a natural question to ask if there are any elements in a group
which commute with all other elements. There are two simple cases to consider:
• If G is Abelian, then every element commutes with every other element!
• The identity e commutes with everything, regardless of the group.
In general, the set of such elements will fall somewhere between these extremes. This subset will
turn out to form a normal subgroup of G.
Definition 13.8. The center of a group G is the subset of G defined by
Z ( G ) := { g ∈ G : ∀h ∈ G, gh = hg}.
We will prove that Z ( G ) / G shortly, although a simple proof (check the axioms of a subgroup
and that the left and right cosets are equal) is straightforward. Instead, here are a few examples.
Examples
1. Z ( G ) = G ⇐⇒ G is Abelian.
2. Z ( D3 ) = {e}. This is straightforward to check as there are only 6 elements!
3. In general Z ( D2n+1 ) = {e} and Z ( D2n ) = {e, ρn/2 }, where ρn/2 is rotation by 180◦ .
For example, it is easy to see in D2n+1 that any rotation and reflection fail to commute.
4. Z (Sn ) = {e} if n ≥ 3.
5. Z (GLn (R)) = {λIn : λ ∈ R× }.
6. Z (On (R)) = {± In }.
The challenge with centers is that they are difficult to compute in general, at least in part becuase
we need to think about non-Abelian groups in order to generate interesting examples. As an example
59
we give an argument for computing Z (GLn (R)).
Let A ∈ GLn (R) be a matrix with only one eigenvector27 v, and let Z ∈ Z (GLn (R)). Then
AZv = ZAv = λZv,
where λ is the sole eigenvector of A. It follows that Zv is an eigenvector of A, whence Zv is parallel
to v. Since our construction depended only on the initial choice of vector v, it follows that Zv must
be parallel to v for every possible choice of v. The only such matrices are multiples of the identity.
Theorem 13.9. Z ( G ) / G.
Proof. We have already done most of the work. Define φ : G → Inn G by φ( g) = c g . Thus each
element φ( g) is the function ‘conjugation by g.’ We quickly see that φ is a homomorphism:
φ( gh)( x ) = c gh ( x ) = ( gh) x ( gh)−1 = g(hxh−1 ) g−1 = c g (ch ( x ))
= φ( g) ◦ φ(h) ( x )
We compute the kernel of φ:
g ∈ ker φ ⇐⇒ c g = id
⇐⇒ ∀h ∈ G, ghg−1 = h
⇐⇒ g ∈ Z ( G )
It follows that ker φ = Z ( G ) / G.
In fact φ : G → Inn G is a surjective homomorphism (every inner automorphism is a conjugation).
The 1st isomorphism theorem tells us that
G
∼
Z ( G ) = Inn G
for any group G.
14
Direct products
Direct products are a straightforward way of defining new groups from old.
Definition 14.1. Let G1 , . . . , Gn be any groups. The direct product of G1 , . . . , Gn is the set
n
∏ Gi = G1 × · · · × Gn
i =1
It is immediate that a direct product has cardinality
n
n
G
∏ i = | G1 | · · · | Gn | = ∏ | Gi |
i =1 i =1
27 Such
A exist: extend v to an orthonormal basis v, v2 , . . . vn of Rn and let A = I + v(v2T + · · · + vnT ). With respect to
this basis, A has 1’s down the main diagonal and the diagonal immediately above the main, and zeros elsewhere: a classic
Jordan form from linear algebra.
60
Example The simplest example is perhaps Z2 × Z2 . Writing Z2 = {0, 1}, we see that
Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}
has four elements. What is interesting is that this set can be given a group structure in an obvious way.
Simply define
( p, q) + (r, s) := ( p +2 r, q +2 s)
and we have a binary operation on Z2 × Z2 . It is straightforward to check that this operation defines
a group. Indeed, we obtain the following Cayley table:
∗
(0, 0)
(0, 1)
(1, 0)
(1, 1)
(0, 0)
(0, 0)
(0, 1)
(1, 0)
(1, 1)
(0, 1)
(0, 1)
(0, 0)
(1, 1)
(1, 0)
(1, 0)
(1, 0)
(1, 1)
(0, 0)
(0, 1)
(1, 1)
(1, 1)
(1, 0)
(0, 1)
(0, 0)
This looks familiar: it is exactly the Cayley table of the Klein 4-group.
Theorem 14.2. Direct products of groups are themselves groups under an elementwise operation: if each Gi
is a multiplictive group, then ∀ ai , bi ∈ Gi define
( a1 , . . . , an ) · (b1 , . . . , bn ) := ( a1 b1 , . . . , an bn )
n
This operation induces a group structure on the direct product ∏ Gi .
i =1
Proof. Simply check the group axioms.
Closure Since each ai bi ∈ Gi this is immediate.
Associativity Each Gi is associative: it is a short calculation to see that the associativity condition
on the whole direct product is equivalent to each individual Gi being associative.
Identity If ei is the identity in Gi , then (e1 , . . . , en ) is the identity in the direct product.
1
Inverse ( a1 , . . . , an )−1 = ( a1−1 , . . . , a−
n ).
The proof uses nothing beyond the fact that each factor Gi is a group in its own right—this should
not be surprising as we have no other information to work with! In particular you should note that
the individual groups Gi do not interact with each other.
Examples
1. Z2 × Z3 = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} under addition (+2 , +3 ). Observe that if we
choose a = (1, 1), then the group can be written (in the above order) as {e, 4a, 2a, 3a, a, 5a}.
Thus Z2 × Z3 is generated by a and is therefore cyclic. Having order 6 we necessarily have
Z2 × Z3 ∼
= Z6 .
2. As observed above, Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}. is isomorphic to the Klein 4-group.
61
3. The direct sum of vector spaces W = U ⊕ V is a more general example. Indeed in linear algebra
it is typical to use direct sum notation rather than Cartesian products. For example the direct
sum of n copies of the real line R is the familiar vector space
Rn =
n
M
i =1
R = R ⊕ · · · ⊕ R.
Orders of elements in direct products In Z12 the element 10 has order 6 = gcd(12
, while in D4
10,12)
the element ρ1 has order 4. What is the order of the element (10, ρ1 ) ∈ Z12 × D4 ? We could simply
compute the cyclic group generated by this element:
n
10,
ρ
)i
=
(0, e), (10, ρ1 ), (8, ρ2 ), (6, ρ3 ), (4, e), (2, ρ1 ),
h(
1
o
(0, ρ2 ), (10, ρ3 ), (8, e), (6, ρ1 ), (4, ρ2 ), (2, ρ3 )
The order is therefore 12. If you consider what happens to each of the entries of each pair, it should
be obvious that the order is the least common multiple of the orders of the originals: 12 = lcm(6, 4). In
general we have the following.
Theorem 14.3. Suppose that ai ∈ Gi has order ri for each i. Then ( a1 , . . . , an ) ∈ G1 × · · · × Gn has order
lcm(r1 , . . . , rn ).
Proof. We simply compute the order:
( a1 , . . . , an )k = ( a1k , . . . , akn ) = (e1 , e2 , . . . , en ) ⇐⇒ ∀i, aik = ei
⇐⇒ ∀i, ri | k
The order is the minimal positive integer k satisfying the above, which is precisely
k = lcm(r1 , . . . , rn )
Example What is the order of28 (1, 4, 3, 2) ∈ Z4 × Z7 × Z5 × Z20 ?
Recalling that the order of x ∈ Zn is gcd(nx,n) we see that the above elements have orders 4, 7, 5 and 10
respectively. Thus the order of (1, 4, 3, 2) is lcm(4, 7, 5, 10) = 140.
Finite(ly generated) Abelian Groups
We have already seen that all finite cyclic groups are isomorphic to some Zn . As we shall see in a
moment, all finite Abelian are fact isomorphic to direct products of these.
We have already seen that Z2 × Z3 ∼
= Z6 but that Z2 × Z2 Z4 . What is the pattern here? Your
gut should suspect that it has something to do with the fact that 2 and 3 are relatively prime.
Theorem 14.4. Zm × Zn ∼
= Zmn ⇐⇒ gcd(m, n) = 1. More generally,
n
∏ Zm
i =1
28 Note
i
∼
= Zm1 ···mn ⇐⇒ gcd(mi , m j ) = 1, ∀i 6= j.
that (1, 4, 3, 2) = ([1]4 , [4]7 , [3]5 , [2]20 ) is a quadruple of equivalence classes!
62
Moreover if n = p1r1 · · · prkk is the unique prime factorization of an integer n, then
Zn ∼
= Z p r1 × · · · × Z p r k
1
k
Since Zm × Zn automatically has order mn the Theorem can be rephrased to say that Zm × Zn is
cyclic ⇐⇒ gcd(m, n) = 1.
Proof. This is merely a corollary of Theorem 14.3. For brevity we just prove the first part.
If gcd(m, n) = 1, then the element (1, 1) ∈ Zm × Zn has order lcm(m, n) = mn. Hence (1, 1) is a
generator of Zm × Zn , which is then cyclic.
Conversely suppose gcd(m, n) = d ≥ 2. Then the maximum order of an element ( p, q) ∈ Zm × Zn is
lcm(m, n) =
mn
mn
=
< mn
gcd(m, n)
d
It follows that Zm × Zn is not cyclic.
Example Is Z5 × Z12 × Z43 cyclic? The Corollary says yes, since no pairs of the numbers 5, 12, 43
have any common factors. It is ghastly to write out, but there are 15 different ways (up to reordering)
of expressing this group!
Z2580 ∼
= Z3 × Z860 ∼
= Z4 × Z645 ∼
= Z5 × Z516 ∼
= Z43 × Z60
∼
∼
∼
= Z12 × Z215 = Z15 × Z172 = Z20 × Z129
∼
= Z3 × Z4 × Z215 ∼
= Z3 × Z5 × Z172
∼
= Z4 × Z5 × Z129 ∼
= Z4 × Z15 × Z43
∼
= Z3 × Z4 × Z5 × Z43
∼
= Z3 × Z20 × Z43
∼
= Z5 × Z12 × Z43
Direct products of cyclic groups have a universal application here. As the next Theorem shows
every finitely generated Abelian group is isomorphic to a direct product of cyclic groups.
Theorem 14.5 (Fundamental Theorem of finitely generated Abelian groups). Every finitely generated
Abelian group is isomorphic to a group of the form
Z pr1 × · · · × Z prnn × Z × · · · × Z,
1
where the pi are (not necessarily distinct) primes, the ri are positive integers and there are a finite number of
factors of Z.
The proof is far too difficult for this course. Its purpose here is to allow us to classify finite
Abelian groups up to isomorphism and we’ll do an example of this below. Definition 7.1 will remind
you what finitely generated means: it is not the same as finite! The above direct product is only a finite
group if it has no factors of Z.
63
Example Find, up to isomorphism, all Abelian groups of order 450. First note that 450 = 2 · 32 · 52 .
Now apply the fundamental theorem to see that the complete list is
1. Z450 ∼
= Z2 × Z32 × Z52
2. Z2 × Z3 × Z3 × Z52
3. Z2 × Z32 × Z5 × Z5
4. Z2 × Z3 × Z3 × Z5 × Z5
Theorem 14.6. If m is a square free integer (@k ∈ Z≥2 such that k2 | m) then there is only one Abelian group
of order m (up to isomorphism).
Proof. By the Fundamental Theorem such a group G must be isomorphic to some Z pr1 × · · · × Z prnn
1
with m = p1r1 · · · prnn . But m being square free implies that every exponenet si is equal to 1 and all the
primes pi are distinct. By Theorem 14.4 we have G ∼
= Z p1 ··· pn = Zm .
All groups of small order As examples of the above we list all the groups of orders 1 through 15
and the Abelian groups of order 16 up to isomorphism. The Fundamental Theorem gives us all the
Abelian groups. In particular observe where Theorem 14.6 applies. There are only two new groups
we haven’t seen before, the Quaternion group Q8 and the Dicyclic group Dic3 (look these up if you’re
interested—these groups are definitely non-examinable!). There are nine non-Abelian groups of order
16 up to isomorphism, six of which we have no notation for in this class! You may be suspicious from
looking at the table that there are no non-Abelian groups of any odd order. This is not so, but you
need to go to order 21 before you find one.
Order
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Abelian
Z1
Z2
Z3
Z4 , V ∼
= Z2 × Z2
Z5
Z6 ∼
= Z2 × Z3
Z7
Z8 , Z2 × Z4 , Z2 × Z2 × Z2
Z9 , Z3 × Z3
Z10 ∼
= Z2 × Z5
Z11
Z12 ∼
= Z3 × Z4 , Z2 × Z6 ∼
= Z2 × Z2 × Z3
Z13
Z14 ∼
= Z2 × Z7
Z15 ∼
= Z3 × Z5
Z16 , Z4 × Z4 , Z2 × Z8 , Z2 × Z2 × Z4 , Z2 × Z2 × Z2 × Z2
64
Non-Abelian
D3 ∼
= S3
D4 , Q8 ∼
= Dic2
D5
D6 , A4 , Dic3
D7
Many
Subgroups of Abelian Groups
Recall Lagrange’s Theorem (10.1) and how it doesn’t have a general converse: if m is a divisr of | G |
then we cannot, in general, claim that G has a subgroup of order m. The smallest group for which this
is evident is A4 which, though it has order 12, has no subgroup of order 6. When G has certain special
forms however, we can find partial converses to Lagrange: for example Corollary 6.8 shows that if G
is cyclic then G has exactly one subgroup for every integer dividing | G |. The following corollaries of
the Fundamental Theorem illustrate that a similar thing happens for finite Abelian groups.
Corollary 14.7 (Existence). Let G be a finite Abelian group. Then G has a subgroup of order m for every
divisor m of | G |.
Proof. By the Fundamental Theorem we can write G ∼
= Z pr1 × · · · × Z prnn . The order of G is then
1
p1r1 · · · prnn . If m is a divisor of | G | then m = p1s1 · · · psnn for some integers si ≤ ri .29 It is straightforward
to check the following:
• pri i −si has order pisi in Z pri .
i
•
D
p1r1 −s1 × · · · × pn
E
rn −sn is a subgroup of G with order m.
There is no uniqueness here however. For example the Klein 4-group Z2 × Z2 has three distinct subgroups of order 2.
Factor Groups and Direct Products
Identifying factor groups of direct products, perhaps using the first isomorphism, is something of a
challenge, bringing together everything in the last few sections.
Examples
1. Classify the factor group (Z4 × Z8 ) h(0, 1)i in terms of the Fundamental Theorem 14.5. Otherwise said, the factor group is a finite Abelian group, so we must be able to write it as a direct
product of Zi ’s.
The cyclic subgroup h(0, 1)i has order 8 and its cosets are
n
o
(Z4 × Z8 )
=
(
0,
0
)
+
0,
1
)i
,
(
1,
0
)
+
0,
1
)i
,
(
2,
0
)
+
0,
1
)i
,
(
3,
0
)
+
0,
1
)i
h(
h(
h(
h(
h(0, 1)i
There are no further cosets (either observe that a non-zero entry in the Z8 part of (i, j) can be
subtracted away by an element in the cyclic group h(0, 1)i, or count elements via the index of
h(0, 1)i; there are (Z4 × Z8 : h(0, 1)i) = 48·8 = 4 cosets).
We therefore have two possibilities: the factor group is isomorphic to Z4 or Z2 × Z2 . Which?
A straightforward answer comes by observing that the factor group is generated by the coset
(1, 0) + h(0, 1)i
29 Even though the primes p do not have to be distinct, we still write m in the above manner and distinguish between
i
exponents si , s j even if pi = p j .
65
whence the factor group is cyclic and thus isomorphic to Z4 . How about obtaining an explicit
isomorphism (Z4 × Z8 ) h(0, 1)i ∼
= Z4 ? Given that the above coset is a generator, simply map
it to a generator of Z4 : indeed
µ : (i, 0) + h(0, 1)i = i
is easily seen to be a well-defined isomorphism.
Comparing with the first isomorphism theorem we see that h(0, 1)i is the kernel of the homomorphism
φ : Z4 × Z8 → Z4 : ( x, y) 7→ x
and that the following diagrams commute.
φ
Z4 × Z8
γ
( x, y) /
8 Z4
'
φ
/9 x
γ
µ
(Z4 × Z8 )
h(0, 1)i
2
'
µ
( x, y) + h(0, 1)i
2. We can do something similar for (Z4 × Z8 ) h(0, 2)i . This time the cyclic subgroup has order
4. As with the previous example, the elements of of the cyclic subgroup only influces second
factor Z8 . We therefore expect to see
Z8
(Z4 × Z8 )
∼
∼
Z
×
=
4
h 2 i = Z4 × Z2 .
h(0, 2)i
This time we try to produce a homomorphism first:
φ : Z4 × Z8 → Z4 × Z2 : ( x, y) 7→ ( x, y mod 2)
This is easily checked to be a well-defined homomorphism. In the context of the first isomorphism theorem we have the following diagrams:
φ
Z4 × Z8
γ
&
/ Z4 × Z2
8
( x, y) φ
γ
µ
(Z4 × Z8 )
h(0, 2)i
/ ( x, y mod 2)
7
%
.
µ
( x, y) + h(0, 2)i
3. The previous examples may have lulled you into a false sense of security: to compute a factor
group you DON’T simply divide by the order of each number in the lower cyclic group. Here
is an example where this approach goes wrong.
You might assume that the factor group (Z4 × Z8 ) h(2, 4)i is isomorphic to
Z4 × Z8
∼
h2i
h 4 i = Z2 × Z4
(∗)
66
but you’d be wrong! Indeed this is very easy to see by counting cosets: the cyclic subgroup
h(2, 4)i has order 2, whence it has
4·8
= 16
Z4 × Z8 : h(2, 4)i =
2
cosets. If (∗) was correct, then’d only have 8 cosets!
The factor group (Z4 × Z8 ) h(2, 4)i is Abelian of order 16. Checking the table on page 64 we
see that there are five possibilities for this group. A little scratch work allows us to identify
which it is:
• Suppose that ( x, y) is an element of the coset ( a, b) + h(2, 4)i. Then ( x − 2, y − 4) is also a
member of the same coset. By considering the first term, this means that there is precisely
one element of each coset whose first entry is 0 or 1.
• It should now be clear that the following elements all lie in distinct cosets of h(2, 4)i:
(0, 0), (0, 1), . . . , (0, 7), (1, 0), . . . , (1, 7).
• It now seems reasonable to conjecture that the factor group is isomorphic to Z2 × Z8 .
Here are two possible ways to prove this:
(a) Observe that the coset (0, 1) + h(2, 4)i has order 8 in the factor group. This narrows our
choice to Z16 or Z2 × Z8 . Now check that all cosets have order at most 8:
8 ( x, y) + h(2, 4)i = (8i, 8j) + h(2, 4)i = (0, 0) + h(2, 4)i
There are no elements of order 16 whene we can rule out Z16 as a candidate. The only
possibility remaining is Z2 × Z8 .
(b) We can define an explicit homomorphism and appeal to the first isomorphism theorem.
φ( x, y) = ( x mod 2, y − 2x mod 8),
φ : Z4 × Z8 → Z2 × Z8 .
Now check the following:
• φ really is a homomorphism: this is obvious since φ is linear in both slots.
φ ( x, y) + (v, w) = φ( x + v, y + w) = x + v mod 2, y + w − 2( x + v) mod 8
= x mod 2, y − 2x mod 8 + v mod 2, w − 2v mod 8
= φ( x, y) + φ(v, w)
• ker φ = h(2, 4)i. For this, note that
φ( x, y) = (0, 0) =⇒ x = 2n for some n ∈ Z
=⇒ y ≡ 4n (mod 8)
=⇒ ( x, y) ∈ h(2, 4)i
Conversely, it is clear that φ(2n, 4n) = (0, 0).
67
• φ is surjective. Observe that
∀( p, q) ∈ Z2 × Z8 , ( p, q) = φ( p, q + 2p)
It follows that (Z4 × Z8 ) h(2, 4)i ∼
= Im φ = Z2 × Z8 .
4. As a final example we do something far more difficult: nothing this tricky is examinable!
Let H = h(4, 2, 3)i be a subgroup of the finitely generated Abelian group Z10 × Z6 × Z. Identify
the factor group G H .
The first problem is counting cosets. Both G and H are infinite groups, so we can’t simply
use the index to find the cardinality of the factor group. However, similarly to the previous
example, the approach is to identify a representative element of each coset. The infinite group
Z in the direct product makes this more challenging so we start with it.
• In every coset ( x, y, z) + H, there is precisely one element ( x, y, z) with z = 0, 1 or 2.
• The elements ( x, y, 0) where x ∈ Z10 and y ∈ Z6 all lie in different cosets: if two such were
in the same coset, then
( x1 , y1 , 0) − ( x2 , y2 , 0) ∈ H ⇐⇒ ( x1 − x2 , y1 − y2 , 0) ∈ H


 x1 ≡ x2 (mod 10)
⇐⇒
and


y1 ≡ y2 (mod 6)
Similarly all the elements ( x, y, 1) and ( x, y, 2) lie in different cosets.
• It follows that the elements ( x, y, 0), ( x, y, 1) and ( x, y, 2) are representatives of different
cosets, and that all cosets have one such representative. There are 10 · 6 · 3 = 180 such
cosets and so the factor group G H is Abelian of order 180.
• The prime decomposition of 180 is 22 · 32 · 5, whence there are, up to isomorphism, four
Abelian groups of order 180, namely
Z4 × Z9 × Z5 ∼
= Z180
Z2 × Z2 × Z9 × Z5 ∼
= Z2 × Z90
Z4 × Z3 × Z3 × Z5 ∼
= Z3 × Z60
Z2 × Z2 × Z3 × Z3 × Z5 ∼
= Z6 × Z30
How do we distinguish between these? Like before we look for elements of a particular
order (lots of scratch work may be required!).
• We compute the order of (1, 1, 1) + H. Certainly its order must be divisible by 3, since the
third entry of
n(1, 1, 1) = (n, n, n) ∈ H
requires 3 | n. Thus let n = 3k. Now
3k (1, 1, 1) + H = (3k, 3k, 3k ) + H = (3k, 3k, 3k ) − k (4, 2, 3) + H
(since (4, 2, 3) ∈ H)
= (−k, k, 0) + H
= H ⇐⇒ 10 | k and 6 | k ⇐⇒ 30 | k ⇐⇒ 90 | n
68
It follows that G H contains an element of order 90. Our choices are narrowed to Z180 and
Z2 × Z90 .
• Finally we show that every element of G H has order dividing 90:
90( x, y, z) + H = (90x, 90y, 90z) − 30z(4, 2, 3) + H = (90x − 120z, 90y − 60z, 0) + H
= (0, 0, 0) + H
There are no elements of order 180, so our factor group cannot be isomorphic to Z180 . We
conclude that G H ∼
= Z2 × Z90 . Phew!
What is perhaps surprising, given our representative elements, is that the factor group is not
isomorphic to Z10 × Z6 × Z3 ∼
= Z6 × Z30 . Given this, finding an explicit surjective homomorphism φ : G → Z2 × Z90 with kernel H seems hard! The trick is to find an element of order 90
(we already have this in (1, 1, 1) + H) and one of order 2 which is not in our order 90 subgroup.
With a little inspection it can be seen that (0, 3, 0) + H is such an element. With a bit of work
we can construct a suitable function: indeed it can be shown that
φ( x, y, z) = x − y mod 2, 27x + 30y + 34z mod 90
is a surjective homomorphism with
ker φ = H,
φ(0, 3, 0) = (1, 0),
φ(1, 1, 1) = (0, 1)
which is inspired by the idea that the corresponding isomorphism µ : G H → Z2 × Z90 satisfies
µ−1 ( X, Y ) = (Y, 3X + Y, Y ) + H
A straightforward calculation confirms that
µ−1 (φ( x, y, z)) = ( x, y, z) + H = γ( x, y, z)
This is all very tricky to work through!
Warning! It may be tempting to take the idea of ‘division of groups’ too far. For example, suppose
we are given that H / G. It might seem reasonable to assume that H × ( G H ) ∼
= G. Unfortunately
life isn’t as simple as this! For example, if we take G = Z8 and H = h4i = C2 , then
G = Z8 ∼
H
C2 = Z4 ,
since the coset 1 + h4i has order 4. But then
Z2 × Z8 C ∼
= Z2 × Z4 Z8 .
2
69
Projections and Subspaces
πi : G1 × · · · Gn → Gi defined as projection onto the ith factor
π i ( g1 , . . . , g n ) = g i
is a homomorphism. Therefore
ker πi = G1 × · · · × Gi−1 × {ei } × Gi+1 × · · · × Gn / G1 × · · · × Gn .
If we are willing to abuse notation we can say that ker πi = G1 × · · · × Gi−1 × Gi+1 × · · · × Gn is a
normal subgroup of the whole direct product. Indeed any combination of projections is a homomorphism (e.g. (π1 , π4 , π5 ) : G1 × · · · × Gn → G1 × G4 × G5 ), and so we can say that
n
∏ Gj / ∏ Gi ,
j∈ I
i =1
for any subset I ⊆ {1, . . . n}.
Let us translate the case when n = 2 using the first isomorphism theorem to identify the factor group
( G1 × G2 ) . The elements of this group are (left) cosets of the form
G1
( g1 , g2 )( G1 × {e2 })
The map µ : ( G1 × G2 ) G → G2 which maps the above coset to g2 is an isomorphism.
1
In the context of vector spaces, this is just the idea that, say, R2 = Span(i, k) is a subspace of R3 ,
viewed as the kernel of the projection map
π2 : ( x, y, z) 7→ (0, y, 0)
We could then claim that
R3 ∼
R = Im π2
R2 =
15
Group actions
For much of these notes, groups have been abstract objects. We have stated theorems in a general
way so that a result may apply as widely as possible. While the approach is worthy and certainly fits
into the general philosophy of pure mathematics, it doesn’t leave one with the feeling that groups
are useful. What is the point of groups, beyond providing a playground for mathematicians?
A simple answer is that groups are useful because of what they do to sets. This idea provided the
first foundations of group theory: recall Cayley’s Theorem that every group is a collection of permutations. But permuting what? This depends on the group and your own creativity: the possibilities
are endless! Indeed we have presented several of our examples in this vein.
• The symmetric group Sn is described in terms of what its elements do to the set {1, . . . , n}.
• The dihedral group Dn is presented as permuting the corners of a regular n-gon.
This is part of a general approach whereby group theory is applicable to the rest of mathematics.
Without the notion of a group acting on a set, group theory is not very useful!
70
Consider the following example as an illustration, where we use group theory to answer a concrete
question: how many genuinely different tetrahedral dice are there?
• We have four faces, and four numbers with which to label them.
It follows that there are 4! = 24 different ways of labelling the
faces of a tetrahedron.
• However, dice roll! We will say that two labellings (i.e. dice) are
identical if and only if one may be rotated into the other. Recall
(page 38) that the rotation group of a regular tetrahedron may be
identified as the alternating group A4 . Therefore, two dice will
be considered indistinguishable if they are related by an element
of the group A4 .
• It seems reasonable, since | A4 | = 12, that the 24 possible dice are
really split into two subsets of 12 dice each. Each subset consists
of 12 dice which may be rotated into each other. There is something to prove here: since we want to argue that a set is partitioned
into distinct subsets, we attempt to define an equivalence relation.
A die x ∈ X
• Let X = {all 24 tetrahedral dice}. An element g ∈ A4 will act on
the set X by rotating all of the dice. If x ∈ X is one die, we write
g · x to denote the new, rotated, die. We define a relation ∼ on X
by
x ∼ y ⇐⇒ ∃ g ∈ A4 such that y = g · x
It is easy to see that this is an equivalence relation on X, and that
there are two equivalence classes, each containing 12 elements.
• It follows that there are really only two distinct 4-sided dice up to
rotations.
The resulting die g · x, if
g = (123)
Of course you don’t need group theory to answer this question. You could argue that any die could
be placed with the number 4 face down, and then rotated so that the number 3 is towards you. There
are then only 2 choices for labelling the other two faces: from left to right 1, 2 or 2, 1. But this is an
easy problem. What if you had a cubic die? Or a 20-sided, icosahedral die? The beauty of the group
theory approach is that it is generalizable. With such a goal in mind, here is the central definition.
Definition 15.1. Let X be a set and G a group. A map · : G × X → X is a (left) action of G on X if,
1. ∀ x ∈ X,
e · x = x,
2. ∀ x ∈ X, ∀ g, h ∈ G,
and,
g · (h · x ) = ( gh) · x.
Otherwise said, the map G → SX : g 7→ g· is a homomorphism.
There is an analogous definition of a right action: in this course, all our actions will be left.
71
Examples
1. Any group G acts on itself by left multiplication. This is Cayley’s Theorem (Thm 8.5).
2. As mentioned earlier, the symmetric Sn group acts on X = {1, 2, . . . , n}.
3. Matrix groups act on vector spaces by matrix multiplication. For example
· : GL2 (R) × R2 → R2 : ( A, v) 7→ Av.
4. A group can act on many different sets. Here are three actions of the orthogonal group O2 (R).
(a) O2 (R) acts on the set X = {1, −1} via A · x := det( A) x.
(b) O2 (R) acts on the set X = R3 via A · v := A(v1 i + v2 j) + v3 k.
(c) O2 (R) acts on the unit circle X = S1 ⊂ R2 via matrix multiplication A · v := Av.
The first two actions of O2 (R) should make you feel somewhat uncomfortable. In the first case,
many matrices A ∈ O2 (R) act in exactly the same way. In essence, the set X is very small relative to
the group: considering the action as a description of the group means that we lose information.
In the second case, it feels like the set X is too large. The group acts via rotations and reflections, but
only rotations around the z-axis, and only reflections across planes containing the z-axis. The action
of O2 (R) doesn’t do anything interesting to vertical vectors.
The third action of O2 (R) on the circle feels nicely balanced:30 the set X is large enough so that the
action fully describes the group, without being inefficiently large. These notions can be formalized.
Definition 15.2. Let G × X → X be an action.
1. Let g ∈ G. The fixed set of g is the set
Fix( g) = { x ∈ X : g · x = x }
(also written Xg )
2. Let x ∈ X. The isotropy subgroup or stabilizer of x is the set
Stab( x ) = { g ∈ G : g · x = x }
(also written Gx )
3. We say that the action is faithful if the only element of g which fixes everything is the identity.
Equivalently
• Fix( g) = X ⇐⇒ g = e.
•
T
x∈X
Stab( x ) = {e}.
4. The action is transitive if one may use the action to transform any element of X to any other:
that is
∀ x, y ∈ X, ∃ g ∈ G such that y = g · x.
30 Perhaps
we should call it a Goldilocks representation.
72
Returning to our examples:
1. The left action of a group on itself is both transitive and faithful.
2. The action of Sn on {1, 2, . . . , n} is both faithful and transitive. For example, for any a, b ∈ X,
the 2-cycle ( a b) maps a 7→ b.
3. The action of matrix multiplication is faithful but not transitive: the zero vector cannot be transformed into any other vector. If we restricted the action of GL2 (R) to the non-zero vectors
X = R2 \ {0}, then the action is both faithful and transitive. To see this, let V, W be matrices
whose first columns are non-zero vectors v, w respectively. Provided that the second column
of each matrix is non-parallel to the first, these are invertible. Since Vi = v and Wi = w, it is
immediate that WV −1 is an invertible matrix whose action maps v 7→ w.
4.
(a) The action A · x = det( A) x is transitive, but not faithful: indeed if A is any orthgonal
matrix with determinant 1, then Fix( A) = X.
(b) The action of O2 (R) on R3 is faithful but not transitive. It is impossible to transform, say,
the zero vector into anything else.
(c) The action of O2 (R) is both faithful and transitive.
Lemma 15.3. As claimed in the definition, Stab( x ) = Gx is a subgroup of G for each x ∈ X.
Proof. Gx is certainly a subset of G, so we must show that it is closed under multiplication and inverses. Let g, h ∈ Gx , then
( gh) · x = g · (h · x ) = g · x = x =⇒ gh ∈ Gx .
Moreover
g · x = x =⇒ g−1 · ( g · x ) = g−1 · x =⇒ ( g−1 g) · x = g−1 · x
=⇒ e · x = g−1 · x =⇒ x = g−1 · x.
Hence g−1 ∈ Gx .
Example The dihedral group D3 acts on the set of corners of an equilateral triangle. Similarly D4
acts on the set of corners of a square. We calculate the stabilizers of each corner under these actions.
See pages 32 and 33 for the descriptions of each element.
Given that the rotations ρ1 , ρ2 move all the corners of the triangle, and
that each reflection µi reflects across the altitude through the corner i, it
is clear that the isotropy subgroup of the corner 1 is
2
Stab(1) = {ρ0 , µ1 }.
1
Similarly Stab(2) = {ρ0 , µ2 } and Stab(3) = {ρ0 , µ3}.
D3 also acts on the set of edges of the triangle X = {1, 2}, {1, 3}, {2, 3} .
Clearly
Stab({1, 2}) = {ρ0 , µ3 },
3
etc.
73
The rotations ρ1 , ρ2 , ρ3 ∈ D4 move all the corners of the square, the reflections µi reflect across the midpoints of edges and δi across diagonals.
The stabilizers come in pairs:
2
3
Stab(1) = Stab(3) = {ρ0 , δ1 },
1
Stab(2) = Stab(4) = {ρ0 , δ2 }.
Acting on the set of edges of the square, we instead obtain
4
Stab({1, 2}) = Stab({3, 4}) = {ρ0 , µ1 },
Stab({1, 4}) = Stab({2, 3}) = {ρ0 , µ2 }.
Orbits
In order to address questions like the dice problem, we need to think about all the elements of a set
X which can be transformed one to another under the action of G.
Definition 15.4. Let G × X → X be an action. The orbit of x ∈ X under G is the set31
Gx = { g · x : g ∈ G } ⊆ X.
Theorem 15.5. The orbits of an action partition X.
Proof. Define x ∼ y ⇐⇒ ∃ g ∈ G such that g · x = y (i.e. x, y are in the same orbit). We claim that ∼
is an equivalence relation.
Reflexivity x = e · x =⇒ x ∼ x.
Symmetry g · x = y =⇒ x = g−1 · y, thus x ∼ y =⇒ y ∼ x.
Transitivity Let y = g · x and z = h · y, then z = (hg) · x, thus x ∼ y and y ∼ z together give
x ∼ z.
Observe that a transitive32 action has only one orbit.
If X is the set {1, 2, . . . n} acted on by a cyclic group G = hσi < Sn , then the definition of orbits
coincides with that given in section 9. Theorem 15.5 is then equivalent to Theorem 9.2.
Theorem 15.6. The cardinality of the orbit containing x ∈ X under the action of G is the index of the isotropy
subgroup Stab( x ):
| Gx | = G : Stab( x ) .
Proof. Observe that
g · x = h · x ⇐⇒ h−1 g · x = x ⇐⇒ h−1 g ∈ Stab( x ) ⇐⇒ g Stab( x ) = h Stab( x )
Thus there are as many distinct elements of the set Gx as there are left cosets of Stab( x ) in G.
31 Be
careful with notation: Gx is the orbit of x under G, while Gx is the stabilizer of x. This is one reason why you might
find it prefereble to write Stab( x ) over Gx .
32 Unhelpfully, we now have two distinct meanings of transitive; one for equivalence relations and one for actions. Take
care!
74
Burnside’s formula
To solve combinatorial problems such as our dice problem, we need to be able to count the number
of orbits of an action. At least when thinking about finite actions, this turns out to be possible in two
different ways.
Theorem 15.7 (Burnside’s formula). Let G be a finite group acting on a finite set X. Then the number of
orbits in X under G satisfies
number of orbits =
1
G
| |
1
∑ |Stab(x)| = |G| ∑ |Fix( g)| .
x∈X
g∈ G
Proof. G and X are finite, whence Theorem 15.6 tells us that the cardinality of the orbit of x is
| Gx | = G : Stab( x ) =
|G|
.
|Stab( x )|
It follows that
1
|G|
∑ |Stab(x)| = ∑
x∈X
x∈X
1
.
| Gx |
(∗)
Now consider a fixed orbit Gy. Clearly | Gx | = | Gy| for each x ∈ Gy, whence
1
| Gy|
=
= 1.
Gx |
| Gy|
x ∈ Gy |
∑
The sum (∗) counts 1 for each distinct orbit in X and therefore returns the number of orbits.
For the second part of the formula, we compute the cardinality of the set
S = {( g, x ) ∈ G × X : g · x = x }.
in two different ways.
• Suppose that x ∈ X is constant. There are |Stab( x )| elements of G which satisfy g · x = x. It
follows that |S| = ∑ Stab( x ).
x∈X
• Now suppose that g ∈ G is constant. There are |Fix( g)| elements of X which satisfy g · x = x.
It follows that |S| = ∑ Fix( g).
g∈ G
We conclude that ∑ |Stab( x )| = ∑ |Fix( g)| and the proof is complete.
x∈X
g∈ G
Examples
1. Consider the set X = {1, 2, 3, 4, 5, 6, 7} under the action of the cyclic group G = h(14)(273)i <
S7 . Since the orbits of a product of disjoint cycles are precisely the cycles, it is clear that wa have
four orbits: {1, 4}, {2, 3, 7}, {5}, {6}. We check Burnside’s formula by computing the stabilizers
and fixed sets. If σ := (14)(273), then | G | = 6. Indeed
σ2 = (237),
σ3 = (14),
σ4 = (273),
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σ5 = (14)(237),
σ6 = e.
The stabilizers and fixed sets are summarized in the following tables
x∈X
Stab( x )
1
{e, σ2 , σ4 }
2
{e, σ3 }
3
{e, σ3 }
4
{e, σ2 , σ4 }
G = {e, σ, σ2 , σ3 , σ4 , σ5 }
5
6
G
{e, σ3 }
7
g∈G
Fix( g)
e
X = {1, 2, 3, 4, 5, 6, 7}
σ
{5, 6}
2
σ
{1, 4, 5, 6}
{2, 3, 5, 6, 7}
σ3
σ4
{1, 4, 5, 6}
σ5
{5, 6}
In either case Burnside just sums the number of elements in all of the subsets in the right column:
4 = number of orbits =
=
1
G
| |
1
|G|
1
∑ |Stab(x)| = 6 (3 + 2 + 2 + 3 + 6 + 6 + 2)
x∈X
1
∑ |Fix( g)| = 6 (7 + 2 + 4 + 5 + 4 + 2)
g∈ G
2. The edges of an equilateral triangle are to be painted using any choice of colors from the rainbow set {red, orange, yellow, green, blue, indigo, violet}. How many distinct triangles are
possible?
This is a little tricky, since we have a lot of choice. We assume that D3 acts on X = {painted
triangles}. Clearly | X | = 73 = 343. It would be very time consuming to compute the stabilizer
of each; fixed sets are easier.
• Fix(e) = X is clear.
• For each of the two rotations ρ1 , ρ2 , the set Fix(ρi ) has 7 elements. If a color-scheme is to be
fixed by ρ1 , then all pairs of adjacent edges must be the same color. The only color-schemes
fixed by ρ1 are therefore those where all sides have the same color.
• For each reflection µi , the set Fix(µi ) has 72 = 49 elements. This is since µi swaps two
edges, which must be the same color in order to be fixed. We have 7 choices for the color
of the pair of switched edges, and an independent choice of 7 colors for the other edge.
It follows that the number of distinct color-schemes is
1
| D3 |
∑
g∈ D3
Fix( g) =
1 3
7
(7 + 7 + 7 + 49 + 49 + 49) = (49 + 1 + 1 + 7 + 7 + 7) = 84
6
6
This question is tricky because we are allowing multiple sides to have the same color. A simpler
question would restrict to the situation where all sides had to be different. In this case D3 acts
on a set of color schemes with cardinality | X | = 7 · 6 · 5 = 210. It is clear that only the identity
has a non-empty fixed set, whence the number of orbits is 210
6 = 35.
3. A cuboid measuring 2 × 2 × 3 cm is to be made into an extremely unfair die by painting the
numbers 1 to 6, one on each face. If we assume that the orientation of the numbers on each face
is irrelevant, how many distinct dice can we make?
76
To apply Burnside we need a set X and group G acting on X such that the orbits of the action
consist of indistinct dice. Thus let X be the set of all possible labelings of the faces and G be the
group of rotations of the prism. There are clearly 6! possible ways to label the faces. The group
of rotations is a little trickier. Consider one of the square faces. We can leave this face where it is
by performing the identity of one of three rotations. Alternatively, we can swap this face with
the other square face before rotating. The result is a rotation group33 of order 8. learly the cyclic
group of order 4, generated by a 90◦ rotation. Every face of a labelled cuboid has a different
number, thus any element of the rotation group changes every labelling, with the exception of
the identity which changes nothing. Therefore
(
X, if g = e,
Fix( g) =
∅, if g 6= e.
It follows that the number of distinct dice is
1
|G|
∑ |Fix( g)| =
g∈ G
6!
= 90.
8
This question can also be answered very simply using combinatorics. There are (62) choices
for the pair of numbers to be painted on the square ends. Once these are chosen, there are
6 configurations of the remaining 4 numbers on the longer sides (up to rotation of the end
squares). We therefore obtain (62) · 6 = 90 distinct dice.
4. Our motivating tetrahedron example be extended to the point where combinatorial arguments
are next to useless. How many dice there are for a given regular polyhedron?
Let X be the set of labelings of the f faces of a polyhedron by the numbers 1, . . . , f . Clearly | X | =
f ! Let G be the group of rotations of the polyhedron. While we many not be able to calculate
this group directly, this doesn’t matter as all we need is the order of the group. Suppose the
polyhedron has f faces which are all the same shape. We may then find f elements g1 . . . , g f
which each map a fixed starting face to one of the f distinct faces. Once there, we can apply
the rotation group of each face. Thus | G | = f | H | where H is the rotation group of the face.
Since the polyhedron is regular, we have H = Cn where n is the number of edges on each face.
Moreover every element of the rotation group except for the identity moves every labeling
(since it moves at least one numbered face). Hence Fix(e) = X and Fix( g) = ∅ for g 6= e. The
number of distinct dice for a regular polyhedron is therefore
f!
( f − 1) !
=
.
fn
n
To summarize:
Polyhedron
Tetrahedron
Cube
Octahedron
Dodecahedron
Icosahedron
f
4
6
8
12
20
n
3
4
3
5
3
# distinct dice
2
30
1,680
7,983,360
40,548,366,802,944,000
33 We don’t need to know G in terms of the standard lists, just its order. However, it isn’t hard to see that the rotation
group is isomorphic to D4 .
77
The same game can be played with non-regular polyhedra. For example suppose you wanted to
make a die out of a football34 constructed from 20 regular hexagons and 12 regular pentagons, with
a different number on each face.
Notice that five hexagons surround each pentagon. Here the rotation
group has order 12 · 5 = 60 (move a pentagon to any other pentagon
then rotate—this is isomorphic to the dodecahedron/icosahedron
33
groups). It is therefore possible to construct 32!
60 = 4.39 × 10 distinct
dice!
Say instead that you had 20 different colored hexagonal patches and 12
different colored pentagonal patches. This time the set of possible footballs X has cardinality 20!12!, so that the number of distinct footballs
25
is 20!12!
60 = 1.94 × 10 .
Just try to do any of the last few calculations using pure combinatorics...
34 A.k.a. a soccer ball for those who’ve never questioned why American football isn’t played with the feet. This is also
the shape of the famous Bucky-ball C60 -molecule with a carbon atom at each corner.
78