Download Notes 1

M3P10/M4P10
Group Theory
John Britnell
ICL 2012
§ 0 Introduction
In this introduction, we review some of the basic definitions and facts about groups, knowledge of which
will be assumed in later chapters. In Section A we recall some results from earlier ICL courses (M1P2 and
M2PM2), and in Section B we list some of the groups with which we will be assuming some familiarity.
This course will be concerned mainly with finite groups, but many facts, both in this chapter and elsewhere, hold for infinite groups too. We shall make it clear when finiteness is necessary for any result.
A. Review of elementary theory
We begin in the natural way, with the definition of a group.
Definition. A group is a pair (G, π), where G is a set, and π is a binary operation G × G −→ G, such that
1.
π(g, π(h, k)) = π(π(g, h), k) for all g, h, k ∈ G (associativity), and
2.
there exists an element e ∈ G such that π(e, g) = π(g, e) = g for all g ∈ G, and with the property that
for every g ∈ G there is an element g−1 ∈ G such that π(g, g−1 ) = π(g−1 , g) = e.
We generally write G for (G, π) when the group operation is unambiguous. We write gh for π(g, h)
(except in cases where there is a standard symbol for the group operation, such as +). In any group G the
element e satisfying the conditions in the second axiom is uniquely determined; we call this element the
identity of G. (We sometimes write eG when we have more than one group in view.) For any g ∈ G, the
element g−1 is also uniquely determined; we call it the inverse of g.
It is not generally true that gh = hg for elements g, h of a group G. A group in which this equality is
satisfied by all elements g and h is said to be abelian.
If G is a group, then a subgroup of G is a non-empty subset of G which is closed under the group
operation, and which contains the inverse of each of its elements. If H is a subgroup of G then we write
H ≤ G. A subgroup of G is itself a group, and has the same identity as G. The intersection of subgroups is a
subgroup; this includes the intersection of infinitely many subgroups. Every group G is a subgroup of itself;
if H ≤ G and H 6= G, then H is said to be a proper subgroup. Every group has a trivial subgroup {e}.
1
Let G be a group, and let g ∈ G. If there exists a positive integer k such that gk = e, then we define the
order of g, written o(g), to be the least such integer. We then have gk = e if and only if k is a multiple of
o(g). If no such integer exists, then we say that g has infinite order. If G is finite then the order of every
element is finite (and in fact divides |G|, by Lagrange’s Theorem, below). In any group, the identity is the
unique element with order 1.
If H ≤ G then a left coset of H in G is any subset of G of the form gH = {gh | h ∈ H}, where g ∈ G.
The cosets of H partition G as a set, i.e. the relation R given by R(g1 , g2 ) ⇐⇒ g1 H = g2 H is an equivalence
relation on G. If H is finite then |gH| = |H| for any g ∈ G. The index of H in G, written |G : H|, is the
number of cosets. If G is finite then |G| = |H| × |G : H|.
z Lagrange’s Theorem
If G is a finite group, and H a subgroup of G, then |H| divides |G|.
Warning. A proper coset of H, i.e. a coset gH which is not equal to H, is never a subgroup of G.
We also have right cosets, of the form Hg = {hg | h ∈ H}. A subgroup N of G is normal if it satisfies
(any of) the following equivalent conditions:
1.
every left coset of N is a right coset;
2.
gN = Ng for every g ∈ G;
3.
gNg−1 = N for every g ∈ G, where gNg−1 = {gng−1 | n ∈ N}.
If N is a normal subgroup of G then we write N E G.
If N E G then we can define a group structure on the set of cosets of N in G by
π(g1 N, g2 N) = g1 g2 N.
The resulting group is called the quotient of G by N, and denoted G/N. There is no such quotient defined
for subgroups which are not normal.
A homomorphism from a group G to a group H is a map ϕ : G −→ H such that ϕ(g1 g2 ) = ϕ(g1 )ϕ(g2 )
for all g1 , g2 ∈ G. The image Im ϕ is a subgroup of H. The kernel Ker ϕ is the set {g ∈ G | ϕ(g) = eH };
the kernel is a normal subgroup of G. If ϕ is both 1–1 (injective) and onto (surjective) then we say that it
is an isomorphism between G and H. (A homomorphism ϕ is injective if and only if Ker ϕ = {e}.) If an
isomorphism between G and H exists then we say that they are isomorphic as groups, and write G ∼
= H.
z First Isomorphism Theorem
Let ϕ : G −→ H be a homomorphism with kernel K E G and image X ≤ H. Then
G/K ∼
= X,
with the isomorphism being given naturally by gK 7→ ϕ(g).
2
For any normal subgroup N of G, the map g 7→ gN defines a surjective homomorphism, called the
canonical map from G to G/N, whose kernel is N. Thus a subgroup of G is normal if and only if it is the
kernel of a homomorphism.
If S ⊆ G, then the subgroup generated by S, written h S i, is the (unique) smallest subgroup containing
S, i.e. the intersection of all subgroups of G containing S. If h S i = G then S is a generating set for G. If
S = {g1 , . . . , gk } then we may write h g1 , . . . , gn i instead of h S i.
If A and B are groups, then the direct product A × B is the group with elements {(a, b) | a ∈ A, b ∈ B}
and multiplication law
(a1 , b1 )(a2 , b2 ) = (a1 a2 , b1 b2 ).
It has order |A||B|, and contains normal subgroups {(a, eB | a ∈ A} and {(eA , b) | b ∈ B} which are isomorphic
to A and B respectively.
B. A selection of groups
B1 Cyclic groups. If G is generated by a single element, G = h g i, then we say that G is cyclic. Cyclic
groups are abelian. Up to isomorphism, there is a unique cyclic group of order n for each positive integer n.
We write Cn for this group when we consider it as an abstract group. There is also a unique infinite cyclic
group C∞ , which is isomorphic to (Z, +), the integers under addition; this group is generated by either of
the elements 1 or −1. We can realize the finite cyclic group of order n concretely as the group of integers
modulo n under addition; this group is (by definition) the quotient of Z by its subgroup nZ = {nk | k ∈ Z}.
If G is any group, and if g is an element of G of order t, then h g i is a cyclic subgroup of order t. Any
subgroup of a cyclic group is cyclic.
B2 Finite abelian groups Any finite abelian group is isomorphic to a direct product of cyclic groups, and are
classified. We give two equivalent forms of the classification; the equivalence comes from the observation
that if G and H are cyclic groups of coprime order, then G × H is cyclic.
z Classification of finite abelian groups
1.
A finite abelian group is isomorphic to a direct product of cyclic groups of prime power order. Two
such groups are isomorphic if and only if they have the same number of direct factors of order pa for each
prime power pa .
For every finite abelian group A, there exists a unique finite sequence of positive integers [a1 , . . . , ak ],
with ak > 1 and such that ai+1 divides ai for all i < k, such that A ∼
= Ca × · · · ×Ca .
2.
1
k
B3 Dihedral groups. The group of symmetries (rotations and reflections) of a plain polygon with n sides,
has order 2n, and is known as the dihedral group. We shall write D2n for this group; but many books prefer
the notation Dn . (So it is a good idea to be explicit when referring to this group: say “symmetry group of an
3
n-gon”, or “dihedral group of order 2n”, if you want to be unambiguous.)
D2n is generated by the set {a, b}, where a is a rotation of the maximal order n, and b is any reflection.
These satisfy the relation bab = a−1 . The cyclic subgroup of order n generated by a is normal in D2n , and
consists of rotations. The rotation subgroup has one proper coset, whose elements are the reflections. All
reflections have order 2.
There is also an infinte dihedral group D∞ , generated by an element a of infinite order, and an element b
of order 2, such that bab = a−1 . This group can be realized as a group of operations on Z, with a : z 7→ z + 1
and b : z 7→ −z for z ∈ Z. We call the elements of the subgroup h a i rotations, and the other elements (which
all have order 2) reflections.
B4 Symmetric groups. Let Ω be any set. A permutation of Ω is a function f : Ω −→ Ω which is invertible
(i.e. it is both injective and surjective). The set of permutations of Ω form a group under composition, called
the symmetric group on Ω.
In the case that Ω = {1, . . . , n} we write Sn for Sym(Ω). The group Sn has order n!. Any element of
Sn may be expressed as a product of disjoint cycles, and this expression is unique up to the order of the
cycles. The notation we commonly use to represent the cycles is shown by the example of (1 2 3 4), which
represents the cycle
1 7→ 2 7→ 3 7→ 4 7→ 1.
This representation becomes unique once the ‘starting point’ of the cycle, 1 in the example, has been chosen.
A cycle with k elements is known as a k-cycle. We often omit the 1-cycles of a permutation (which contain
the fixed points) when we write it in disjoint cycle notation. The order of a permutation is the lowest common
multiple of its cycle lengths.
The dihedral group D2n arises naturally as a subgroup of Sn ; we obtain it by numbering the vertices
(corners) of an n-gon with the numbers 1, . . . , n, and then representing each symmetry of the n-gon by the
permutation that it induces on the vertices. (By choosing different vertex labellings we get distinct, but
isomorphic, subgroups of Sn .)
The symmetric group Sn−1 arises naturally as the subgroup of Sn , consisting of all permutations which
fix the point n. (By choosing a different point to fix we get a distinct, but isomorphic, subgroup.)
B5 Alternating groups. A transposition on Ω is a permutation which interchanges two points of Ω, fixing
the rest. A permutation is said to be even if it is a product of an even number of transpositions, and odd if it
is a product of an odd number.
z Parity Theorem
Each element of Sn is even or odd, but no element is both even and odd.
4
The even permutations of Sn form a subgroup of index 2, known as the alternating group An . The order of
the group An is n!/2 for n > 1. A permutation with exactly one non-trivial cycle lies in An if and only if that
cycle has odd length. In general, a permutation expressed in disjoint cycle notation belongs to An if and only
if it has an even number of cycles of even length.
B6 Small groups. We list the groups with order 8 or less, with which we shall assume familiarity later.
The trivial group. There is (obviously) only one group of order 1 (up to isomorphism). This group we shall
generally refer to as {e} or simply 1.
Groups of prime order. If G has prime order p then any non-identity element has order p, and so G is cyclic.
Thus there is a unique group of order p for each prime p (up to isomorphism). This accounts for groups of
orders 2, 3, 5, 7.
Groups of order 4. A group of order 4 is either cyclic, or else isomorphic to the Klein 4 group V4 . This has
non-identity elements a, b, c of order 2, with ab = ba = c, bc = cb = a and ca = ac = b. The group V4 is
abelian. It occurs as a normal subgroup of S4 , whose non-identity elements are the double transpositions,
(1 2)(3 4), (1 3)(2 4) and (1 4)(2 3).
Groups of order 6. Any group of order 6 is either cyclic, or else isomorphic to S3 . (In particular, D6 ∼
= S3 .)
Groups of order 8. There are three abelian groups of order 8, namely C8 , and C4 ×C2 , and C2 ×C2 ×C2 .
There are also two non-abelian groups of order 8, namely the dihedral group D8 , and the quaternionic
group Q8 :
Q8 = {1, −1, i, −i, j, −j, k, −k},
where 1 is the identity, and the group operation satisfies
i(−i) = j(−j) = k(−k) = 1,
ij = k,
ji = −k.
(Given that Q8 is a group, the full multiplication table is determined by these conditions.)
§ 1 More on Quotient Groups
The course properly begins here. We start by looking in more detail at homomorphisms on a group G,
and in particular, at how they behave on subgroups of G. Throughout this chapter, we shall suppose that
θ is a homomorphism from G onto a group H, whose kernel is K E G. We shall usually suppose that θ is
surjective; in this case the First Isomorphism Theorem tells us that
G∼
= H.
K
5
For a subgroup L ≤ G, we define θ (L) = {θ (`) | ` ∈ L}. Conversely, for a subgroup X ≤ H, we define
θ −1 (X) = {g ∈ G | θ (g) ∈ X}.
Proposition 1. Let θ : G −→ H be surjective, with kernel K. Then
1.
θ (L) is a subgroup of H, for every subgroup L of G.
2.
θ −1 (X) is a subgroup of G, for every subgroup X fo H.
3.
θ (θ −1 (X)) = X for every subgroup X of H.
4.
θ −1 (θ (L)) = L for every subgroup L of G such that K ≤ L.
Proof.
1.
If x, y ∈ θ (L), then there exist `, m ∈ L with θ (`) = x, θ (m) = y. Now θ (`m) = xy and θ (`−1 ) = x−1 .
2.
If `, m ∈ θ −1 (X) then θ (`m) = θ (`)θ (m) ∈ X, and θ (`−1 ) = θ (l)−1 ∈ X.
3.
It is obvious that θ (θ −1 (X)) ⊆ X; and since X ≤ Im θ , the reverse containment is clear too.
4.
It is obvious that L ⊆ θ −1 (θ (L)). If g ∈ θ −1 (θ (L)), then there exists ` ∈ L such that θ (g) = θ (`). We
have g`−1 ∈ K. Now if K ≤ L, then g`−1 ∈ L, and hence g ∈ L, as required.
Example. Let G = (Z, +), and let θ be the canonical map onto the quotient Z/nZ, where n ∈ N. This
quotient is cyclic of order n, and it has a subgroup dZ/nZ for each divisor d of n. It is clear that θ (dZ) =
dZ/nZ, and that θ −1 (dZ/nZ) = dZ, for each divisor d. If m is any integer, then θ (mZ) = aZ/nZ, where
a = hcf(m, n), and then θ −1 (θ (mZ)) = aZ. Thus we have θ −1 (θ (mZ)) = mZ only if hcf(m, n) = m, i.e.
only if m divides n; this is equivalent to the condition that mZ contains nZ, the kernel of θ .
Proposition 2. Let θ : G −→ H be a surjective homomorphism with kernel K. The map L 7→ θ (L) gives a
bijection between subgroups of G containing K, and subgroups of H. If K ≤ L, then L E G ⇐⇒ θ (L) E H.
Proof. Let L, M be subgroups of G containing K. If θ (L) = θ (M) then θ −1 (θ (L)) = θ −1 (θ (L)), and hence
L = M by Proposition 1.4. Let X ≤ H; then θ (θ −1 (X)) = X by Proposition 1.3, and clearly θ −1 (X) is a
subgroup of G containing K. Thus the map L 7→ θ (L) is a bijection between subgroups of G containing K,
and subgroups of H, as required. Suppose that L E G, and let X = θ (L). Let H ∈ H; since θ is surjective
there exists g ∈ G with θ (g) = h. Now gLg−1 = L, and hence hXh−1 = θ (gLg−1 ) = θ (L) = X; so X E H.
Example. Recall that S4 has a normal subgroup V4 = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. The quotient
S4 /V4 has no elements of order 6 (since the largest order of an element of S4 itself is 4), and so it is not
cyclic; therefore it is isomorphic to S3 . Let θ : S4 −→ S3 be a homomorphism with kernel V4 .
So by Proposition 2, we can put the subgroups of S4 containing V4 into bijection with the subgroups
of S3 . Some of the corresponding pairs are clear: θ (S4 ) = S3 , and the since V4 is the kernel we have
θ (V4 ) = {e}. The subgroup A4 ≤ S4 contains an element of order 3; this is not in the kernel, and so must
6
map to an element of order 3 in S3 . It follows that θ (A4 ) = A3 , since A3 is the only proper subgroup of S3
containing a 3-cycle.
The three remaining subgroups of S3 all have order 2. The corresponding subgroups of S4 have order
properly divisible by 4 (since they properly contain V4 ), but not 12 or 24 (since the only subgroups of these
orders are A4 and S4 ). Hence they must have order 8. In fact these are the dihedral groups D8 , obtained as
subgroups of S4 by labelling the vertices of a square 1, 2, 3, 4 and considering the resulting permutations.
(There are three essentially different labellings, giving three distinct subgroups of S4 .)
Proposition 3. Let θ : G −→ H have kernel K. Let L be a subgroup of G such that K ≤ L, and let X = θ (L).
Then K E L, and L/K ∼
= X.
Proof. Let ` ∈ L. Since ` ∈ G, and since K E G, we have `K = K`; hence K E L. Let θL be the restriction
of θ to L; then Ker θL = K and Im θL = X, and so L/K ∼
= X by the First Isomorphism Theorem.
Theorem 4. z Second Isomorphism Theorem
Let L be a normal subgroup of G, and let K be a subgroup of L such that K E G. Then
G/K ∼ G
= .
L/K
L
Proof. Define a map θ : G/K −→ G/L by θ : gK 7→ gL. Note that this map is well defined, since if hK = gK
then h−1 g ∈ K, and hence h−1 g ∈ L and hL = gL. Now θ is a homomorphism, since for all g, h ∈ G we have
θ ((gK)(hK)) = θ (ghK) = ghL = (gL)(hL) = θ (gK)θ (hK).
We observe that θ (gK) = L ⇐⇒ gL = L ⇐⇒ g ∈ L, and so Ker θ = L/K. It is clear that Im θ = G/L, and
the result now follows by the First Isomorphism Theorem.
We have now dealt fully with the behaviour of θ on those subgroups of G which contain the kernel of θ .
It remains to say what we can about a general subgroup of G. The following proposition will be be needed
here, and later in the course. For subsets S and T of a group G, we define ST := {st | s ∈ S,t ∈ T }.
Proposition 5. Let A and B be subgroups of a group G. Then
1.
AB is a subgroup of G if and only if AB = BA.
2.
If A and B are finite then |AB| = |A||B|/|A ∩ B|.
Proof. [This was omitted in lectures, being the solution to Homework 1, Qu. 1.]
1.
We show that the condition AB = BA is equivalent to the condition that AB contains inverses for all
of its elements. Suppose a ∈ A and b ∈ B. Then (ba)−1 = a−1 b−1 ∈ AB. Thus for g ∈ G we have g ∈
BA ⇐⇒ g−1 ∈ AB. Now it is clear that the condition g ∈ AB ⇐⇒ g−1 ∈ AB is equivalent to the condition
g ∈ AB ⇐⇒ g ∈ BA, i.e. AB = BA, as claimed. This is enough to show that if AB is a subgroup then
7
AB = BA; to show the converse, we just need to check closure under multiplication in the case that AB = BA.
For a1 , a2 ∈ A and b1 , b2 ∈ B we have (a1 b1 )(a2 b2 ) ∈ a1 (BA)b2 = a1 (AB)b2 = AB, as required.
2.
Notice that |A × B| = |A||B|. Define a map f : A × B =⇒ AB by f : (a, b) 7→ ab. Now we have
f (a1 , b1 ) = f (a, b) ⇐⇒ a1 b1 = ab ⇐⇒ a−1 a1 = bb−1
1 .
But this implies that a−1 a1 ∈ A ∩ B. It is clear from this that for a ∈ A we can find b ∈ B with ab = a1 b1 if
and only if a−1 a1 ∈ A ∩ B, and hence there are |A ∩ B| elements of A × B which f maps to a1 b1 . Thus f is a
|A ∩ B|-to-1 map, and the result follows.
Proposition 6. Let K be a normal subgroup, and L a subgroup, of a group G. Then
1.
KL is a subgroup of G. It is the smallest subgroup containing K and L.
2.
K E KL.
3.
If θ is a homomorphism on G with kernel K, then θ (L) = θ (KL).
Proof.
1.
Since K is normal, `K = K` for all ` ∈ L, and so KL = LK. Now KL is a subgroup by Proposition
5.1. Clearly it contains K = Ke and L = eL; furthermore any subgroup containing both K and L contains all
elements k` for k ∈ K and ` ∈ L, and so contains KL; therefore KL is the smallest subgroup containing K
and L.
2.
This is clear, since K E G.
3.
It is obvious that θ (K) ⊆ θ (KL). But if k ∈ K and ` ∈ L then θ (k`) = θ (k)θ (`) = θ (`) (since K =
Ker θ )), and hence θ (KL) ⊆ θ (L).
Theorem 7. z Third Isomorphism Theorem
Let K be a normal subgroup, and L a subgroup, of a group G. Then K ∩ L is a normal subgroup of L, and
KL ∼ L
.
=
K
K ∩L
Proof. Let θ be the canonical homomorphism from G onto the quotient G/K, and let θL be the restriction
of this map to L. Then θL is a homomorphism. It is clear that Ker θL = Ker θ ∩ L = K ∩ L (from which it
follows that K ∩ L E L). Since L ≤ KL it is obvious that Im θL ≤ KL/K. But if g ∈ KL then (since KL = LK)
we may write g = `k for k ∈ K and ` ∈ L, and now gK = `kK = `K = θL (`). So Im θL = KL/K, and the
result now follows by the First Isomorphism Theorem.
Warning. The numbering of the Group Isomorphism Theorems is not standardized. Many sources switch
the order of the Second and Third Theorems as given here. Other sources have only two theorems, or else
four. The only completely unambiguous way to cite one of these theorems is to state it.
8