Download A group homomorphism is a function between two groups that links

INTRODUCTION TO GROUP THEORY (MATH 10005)
51
15. H OMOMORPHISMS AND NORMAL SUBGROUPS
A group homomorphism is a function between two groups that links the two
group operations in the following way.
Definition 15.1. Let (G, ∗) and (H, •) be groups. A group homomorphism ϕ : G → H
is a function such that
ϕ(x ∗ y) = ϕ(x) • ϕ(y)
for all x, y ∈ G.
Example 15.1. An isomorphism is a homomorphism. The difference is that we don’t
require a homomorphism to be bijective. If it is, then it is an isomorphism.
Remark 15.1. If the groups G and H are written multiplicatively, then ϕ : G → H is a
homomorphism if
ϕ(xy) = ϕ(x)ϕ(y)
for all x, y ∈ G. But it should be noted that on the left hand side of that equation we
multiply x and y in G, but on the right hand side we multiply ϕ(x) and ϕ(y) in H, so
this still links the group operations of different groups.
Remark 15.2. If you are doing Linear Algebra, then you may find it helpful to note
a similarity between the definitions of a group homomorphism and a linear map.
These are both functions that “commute with the operations” in the sense that the
definition of a homomorphism says that multiplying two elements and then applying the homomorphism gives the same as applying the homomorphism to the two
elements and then multiplying the resultan elements, and the definition of a linear
map says the same for the operations of addition and multiplication by scalars in
place of the group operation. In fact, many of the basic facts about homomorphisms
are very similar to basic facts about linear maps, usually with pretty much the same
proof.
Example 15.2. If G and H are groups and eH is the identity element of H, the function
ϕ : G → H given by
ϕ(x) = eH
for all x ∈ G is a homomorphism (the trivial homomorphism).
Example 15.3. If H ≤ G are groups, then the inclusion map i : H → G given by
i(x) = x ∈ G
for all x ∈ H is a homomorphism. This is injective but not surjective (unless H = G).
52
COURSE NOTES 2016-2017
The existence of an isomorphism between two groups implies that the two groups
have the same abstract structure, as we have seen. The existence of a homomorphism doesn’t: indeed, the example of the trivial homomorphism shows that there is
a homomorphism between any two groups. However, (non-trivial) homomorphisms
from a group G to a more familiar group H can still be very useful tools to understand
G.
Proposition 15.1. Let H and K be groups, and let H × K be the direct product. Then there
are homomorphisms
(1)
(2)
(3)
(4)
iH : H → H × K with iH (x) = (x, eK ) for x ∈ H.
ik : K → H × K with iK (y) = (eH , y) for y ∈ K.
πH : H × K → H with πH (x, y) = x for x ∈ H, y ∈ K.
πK : H × K → K with πK (x, y) = y for x ∈ H, y ∈ K.
Proof. iH is a homomorphism since, for x, x 0 ∈ H,
iH (xx 0 ) = (xx 0 , eK ) = (x, eK )(x 0 , eK ) = iH (x)iH (x 0 ).
πH is a homomorphism since, for x, x 0 ∈ H and yy, y 0 ∈ K,
πH ((x, y)(x 0 , y 0 )) = πH (xx 0 , yy 0 ) = xx 0 = πH (x, y)πH (x 0 , y 0 ).
The proofs for iK and πK are similar.
Lemma 15.2. Let ϕ : G → H be a homomorphism, let eG and eH be the identity elements of
G and H respectively, and let x ∈ G. Then
(1) ϕ(eG ) = eH ,
(2) ϕ(x−1 ) = ϕ(x)−1 ,
(3) ϕ(xi ) = ϕ(x)i for any i ∈ Z.
Proof.
(1) ϕ(eG ) = ϕ(eG eG ) = ϕ(eG )ϕ(eG ), so by uniqueness of the identity ϕ(eG ) =
eH .
(2) eH = ϕ(eG ) = ϕ(xx−1 ) = ϕ(x)ϕ(x−1 ), so by uniqueness of inverses ϕ(x−1 ) =
ϕ(x)−1 .
(3) This is true for positive i by a simple induction (it is true for i = 1, and if it is
true for i = k then
ϕ(xk+1 ) = ϕ(xk )ϕ(x) = ϕ(x)k ϕ(x) = ϕ(x)k+1 ,
so it is true for i = k + 1). Then by (2) it is true for negative i, and by (1) it is
true for i = 0.
Definition 15.2. Let ϕ : G → H be a homomorphism and eH the identity element of
H. The kernel of ϕ is
ker(ϕ) = {x ∈ G : ϕ(x) = eH } ⊆ G,
INTRODUCTION TO GROUP THEORY (MATH 10005)
53
and the image of ϕ is
im(ϕ) = {ϕ(x) : x ∈ G} ⊆ H.
Theorem 15.3. If ϕ : G → H is a homomorphism, then ker(ϕ) is a subgroup of G and
im(ϕ) is a subgroup of H.
Proof. Since, by Lemma 15.2, ϕ(eG ) = eH , eG ∈ ker(ϕ).
If x, x 0 ∈ ker(ϕ) then
ϕ(xx 0 ) = ϕ(x)ϕ(x 0 ) = eH eH = eH ,
so xx 0 ∈ ker(ϕ), and so ker(ϕ) is closed under multiplication.
−1
If x ∈ ker(ϕ) then by Lemma 15.2 ϕ(x−1 ) = ϕ(x)−1 = e−1
∈ ker(ϕ),
H = eH , so x
and so ker(ϕ) is closed under taking inverses.
So ker(ϕ) is a subgroup of G.
eH = ϕ(eG ) by Lemma 15.2, so eH ∈ im(ϕ).
If y, y 0 ∈ im(ϕ) then y = ϕ(x) and y 0 = ϕ(x 0 ) for some x, x 0 ∈ G, and so
yy 0 = ϕ(x)ϕ(x 0 ) = ϕ(xx 0 )
is in im(ϕ), and so im(ϕ) is closed under multiplication.
If y ∈ im(ϕ) then y = ϕ(x) for some x ∈ G, and by Lemma 15.2
y−1 = ϕ(x)−1 = ϕ(x−1 ),
so y−1 ∈ im(ϕ), and so im(ϕ) is closed under taking inverses.
So im(ϕ) is a subgroup of H.
Theorem 15.4. Let ϕ : G → H be a homomorphism. Then ϕ is injective if and only if
ker(ϕ) = {eG }.
Proof. If ϕ is injective, then there is at most one element x ∈ G with ϕ(x) = eH . But
ϕ(eG ) = eH , so ker(ϕ) = {eG }.
If ϕ is not injective, then there are two elements x 6= y of G with ϕ(x) = ϕ(y). But
then ϕ(xy−1 ) = ϕ(x)ϕ(y)−1 = eH , so eG 6= xy−1 ∈ ker(ϕ), and so ker(ϕ) 6= {eG }.
Example 15.4. Let D2n be the dihedral group of order n, let H = ({1, −1}, ×) and define
ϕ : D2n → H by

1
if x is a rotation
ϕ(x) =
−1 if x is a reflection.
Then it is easy to check that ϕ is a homomorphism. ker(ϕ) = hai is the subgroup
containing all rotations, and of course ϕ is surjective, so im(ϕ) = H.
54
COURSE NOTES 2016-2017
Example 15.5. Let Sn be the symmetric group of degree n, let H = ({1, −1}, ×) and
define ϕ : Sn → H by

1
if x is an even permutation
ϕ(x) =
−1 if x is an odd permutation.
Then it is easy to check that ϕ is a homomorphism. ker(ϕ) is the alternating group
An , and if n > 1 then ϕ is surjective, and so im(ϕ) = H.
Example 15.6. Example 15.4 shows that hai is the kernel of a homomorphism from
D2n to another group. What about other subgroups, such as hbi?
Suppose ϕ : D2n → H is a homomorphism with b ∈ ker ϕ. Then
ϕ(aba−1 ) = ϕ(a)ϕ(b)ϕ(a)−1 = ϕ(a)eH ϕ(a)−1 = eH ,
so aba−1 , which is equal to a2 b, must also be in ker(ϕ). So if n > 2 (so that a2 b 6= b)
then hbi cannot be the kernel of a homomorphism. This motivates the following
definition.
Definition 15.3. Let G be a group. A normal subgroup of G is a subgroup N ≤ G
such that gxg−1 ∈ N for every g ∈ G and x ∈ N. The notation N E G is used to mean
that N is a normal subgroup of G, or N C G if we want to specify that N 6= G.
We can generalize the argument in the last example:
Theorem 15.5. Let ϕ : G → H be a homomorphism. Then ker(ϕ) E G.
Proof. We have already proved that ker(ϕ) is a subgroup of G.
Let g ∈ G and x ∈ ker(ϕ). Then
ϕ(gxg−1 ) = ϕ(g)ϕ(x)ϕ(g)−1 = ϕ(g)eH ϕ(g)−1 = eH ,
so gxg−1 ∈ ker(ϕ), and so ker(ϕ) is a normal subgroup of G.
Remark 15.3. In fact, finding a homomorphism ϕ such that N = ker(ϕ) is often the
easiest way to prove that N is a normal subgroup of G.
Example 15.7. H = hbi is not a normal subgroup of D2n if n > 2, since aba−1 6∈ H.
However, K = hai is a normal subgroup since we showed in Example 15.4 that it is
the kernel of a homomorphism.
Similarly, the alternating group An is a normal subgroup of Sn , as we showed in
Example 15.5 that it is the kernel of a homomorphism.
Proposition 15.6. If G is an abelian group then every subgroup of G is normal.
Proof. Let N ⊆ G, g ∈ G and x ∈ N. Then if G is abelian, gxg−1 = x ∈ N, and so
N E G.
INTRODUCTION TO GROUP THEORY (MATH 10005)
55
There are some other ways to formulate the definition of a normal subgroup. Recall the definition of the left coset gN = {gx : x ∈ N} and right coset Ng = {xg : x ∈ N}
for N ≤ G and g ∈ G. Using similar notation we define
gNg−1 = gxg−1 : x ∈ N .
Proposition 15.7. Let G be a group and N ≤ G. The following are equivalent.
(1) N E G.
(2) gNg−1 = N for all g ∈ G.
(3) gN = Ng for all g ∈ G.
Proof. (1) ⇒ (2): Suppose N E G and g ∈ G. The definition of “normal subgroup”
says that gNg−1 ⊆ N for all g ∈ G. But applying this to g−1 this in particular implies
that g−1 Ng ⊆ N and so
x ∈ N ⇒ g−1 xg ∈ N ⇒ x = g(g−1 xg)g−1 ∈ gNg−1 .
(2) ⇒ (1) is trivial.
If gNg−1 = N then for gx ∈ gN, gx = (gxg−1 )g ∈ Ng so gN ⊆ Ng and similarly
Ng ⊆ gN, so (2) ⇒ (3).
If gN = Ng and x ∈ N, then gx ∈ Ng, so gx = yg for some y ∈ N, and so
gxg−1 = ygg−1 = y ∈ N, so gNg−1 ⊆ N, so (3) ⇒ (1).
Remark 15.4. This tells us that a subgroup is normal if and only if its left cosets are the
same as its right cosets. Let’s look at how this works in the example of N = hai E D6
and the left and right cosets bN and Nb.
Since N = {e, a, a2 },
bN = {be, ba, ba2 } = {b, ab2 , ab}
and
Nb = {eb, ab, a2 b}.
So as expected since N E D6 , bN = Nb.
However, notice that this is NOT saying that bx = xb for every x ∈ N. Indeed, in
this example ab 6= ba = a2 b. For x = a it is true that xb is in bN, but it is equal to by
for an element y ∈ N that is different from x.
56
COURSE NOTES 2016-2017
16. Q UOTIENT GROUPS
We’ve seen that the kernel of a homomorphism is always a normal subgroup. Now
we’ll look at an important construction that shows that every normal subgroup is the
kernel of some homomorphism.
Let’s start with a familiar example, (Z/nZ, +). We can think of this as being constructed from the group (Z, +) and the subgroup nZ = {ns : s ∈ Z} as follows:
We regard elements of Z (i.e., integers) as “equivalent” if their difference is in nZ
(i.e., if they are congruent (mod n)). Then we form a new group Z/nZ whose elements are equivalence classes [s] of elements of Z and with group operation [s][t] =
[s + t], which we check is well-defined.
In fact, we can do exactly the same construction for any abelian group G and subgroup H, but if G is not abelian, then we have to be more careful about what the
“difference” between two elements x and y means: do we mean xy−1 or y−1 x, which
may be different? This is why it turns out to be important that our subgroup is normal.
Let’s look at a simple example, to point out the problems.
Example 16.1.
• Let G = D6 and let N be the (normal) subgroup hai. Then there
are two left cosets N (consisting of the rotations) and bN (consisting of the
reflections). If we compose two rotations we get a rotation; if we compose two
reflections we get a rotation, and if we compose (in either order) a rotation and
a reflection then we get a reflection. So if we want to know which coset xy is
in, then we only need to know which cosets x and y are in.
• Let G = D6 and let H be the (non-normal) subgroup hbi. Then there are
three left cosets H = {e, b}, aH = {a, ab} and a2 H = {a2 , a2 b}. Observe what
happens if we multiply two elements x ∈ H and y ∈ aH: if x = e and y = a
then xy = a, which is in the coset aH, but if x = b and y = ab then xy =
bab = a2 , which is in the coset a2 H. So the coset that contains xy does not
depend only on which cosets contain x and y.
Let G be a group and N E G a normal subgroup. Remember that this means that
the left coset xN and right coset Nx are the same for every x ∈ N. To simplify the
notation, and point out the analogy with modular arithmetic, we’ll use the notation
[x] = xN = Nx
for the cosets.
Lemma 16.1. If x, y ∈ G, then [xy] depends only on [x] and [y].
Proof. Let xn ∈ [x] and yn 0 ∈ [y]. Then xnyn 0 = xy(y−1 ny)n 0 , which is in xyN since
N is normal and so y−1 ny ∈ N.
INTRODUCTION TO GROUP THEORY (MATH 10005)
57
Definition 16.1. If G is a group and N E G a normal subgroup. The quotient group
G/N is the set of cosets of N in G, with the binary operation (written multiplicatively)
[x][y] = [xy],
for x, y ∈ G.
By the previous lemma, this binary operation is well-defined. But of course we
need to check that this is a group.
Theorem 16.2. If N E G, then G/N is a group.
Proof. Since
([x][y])[z]) = [xy][z] = [xyz] = [x][yz] = [x]([y][z]),
for any x, y, z ∈ G, it is associative.
Since
[e][x] = [ex] = [x] = [xe] = [x][e]
for any x ∈ G, [e] is an identity element.
Since
[x][x−1 ] = [xx−1 ] = [e] = [x−1 x] = [x−1 ][x]
for any x ∈ G, [x−1 ] is an inverse to [x].
There is a natural function π : G → G/N defined by sending an element of G to the
coset that contains it: π(x) = [x].
Proposition 16.3. π : G → G/N is a surjective homomorphism with ker(π) = N.
Proof. Clearly π is surjective, since every element of G/N is of the form [x] = π(x) for
some x ∈ G.
If x, y ∈ G then
π(xy) = [xy] = [x][y] = π(x)π(y),
so π is a homomorphism.
Since, for x ∈ G,
x ∈ ker(π) ⇔ [x] = [e] = N ⇔ x ∈ N,
ker(π) = N.
In fact, we’ll see later that this Proposition characterizes G/N up to isomorphism.
It’s really because of this property that quotient groups are useful and important,
and the details of the construction using cosets are only important in order to prove
that there is a group G/N with this property.
58
COURSE NOTES 2016-2017
17. T HE H OMOMORPHISM T HEOREM
In this section we’ll prove an important theorem that gives a link between the
kernel and image of a group homomorphism. This is variously known as the “Homomorphism Theorem” or the “First Isomorphism Theorem” (although the second
name is also sometimes used for a different theorem).
Recall that if ϕ : G → H is a group homomorphism, then ker(ϕ) is a normal
subgroup of G, and so it makes sense to form the quotient group G/ ker(ϕ).
Theorem 17.1 (Homomorphism Theorem). Let ϕ : G → H be a group homomorphism.
Then the quotient group G/ ker(ϕ) is isomorphic to im(ϕ).
Proof. We’ll prove this by constructing an isomorphism
α : G/ ker(ϕ) → im(ϕ).
Using the notation from the previous section, where if x ∈ G then [x] ∈ G/ ker(ϕ)
denotes the coset x ker(ϕ), let
α([x]) = ϕ(x).
There are a number of things that we need to check. First, that α is well-defined.
Suppose [x] = [x 0 ]. Then x ker(ϕ) = x 0 ker(ϕ), and so x = xe = x 0 k for some
k ∈ ker(ϕ). But then
ϕ(x) = ϕ(x 0 k) = ϕ(x 0 )ϕ(k) = ϕ(x 0 )e = ϕ(x 0 ),
and so α is well-defined, as it doesn’t make any difference which of the elements
x, x 0 ∈ [x] we use.
Next, let’s prove that α is a homomorphism. This is true, since if [x], [y] ∈ G/ ker(ϕ)
then
α([x][y]) = α([xy]) = ϕ(xy) = ϕ(x)ϕ(y) = α([x])α([y]).
Now we just need to show that α is bijective. It is injective since
α([x]) = α([y]) ⇒ ϕ(x) = ϕ(y)
⇒ ϕ(x−1 y) = e
⇒ y = x(x−1 y) ∈ x ker(ϕ)
⇒ [x] = [y],
since if two left cosets [x] = x ker(ϕ) and [y] = y ker(ϕ) have an element (y in this
case) in common, then they are equal.
Finally, it is easy to see that α is surjective, since if h ∈ im(ϕ) then
h = ϕ(g) = α([g])
for some g ∈ G.
INTRODUCTION TO GROUP THEORY (MATH 10005)
59
This means that finding a homomorphism between two groups immediately gives
us lots of information about the kernel and image. To summarize:
Theorem 17.2. Let ϕ : G → H be a homomorphism. Then ker ϕ is a normal subgroup of
∼ G/ ker(ϕ).
G, im(ϕ) is a subgroup of H, and im(ϕ) =
Example 17.1. Let Sn be the symmetric group of degree n, where n > 1. Let ϕ : Sn →
(R \ {0}, ×) be the homomorphism with ϕ(σ) = 1 if σ is an even permutation and
ϕ(σ) = −1 if σ is an odd permutation. Then the kernel of ϕ is An and the image is
{1, −1}, so An E Sn and
∼ ({1, −1}, ×) ,
Sn /An =
which is a cyclic group of order 2.
Example 17.2. Let D2n be the dihedral group of order 2n, and let ϕ : D2n → (R \ {0}, ×)
be the homomorphism with ϕ(x) = 1 if x is a rotation and ϕ(x) = −1 if x is a
reflection. Then the kernel of ϕ is the subgroup hai < D2n , which is therefore a
normal subgroup, and the image of ϕ is {1, −1}, so
∼ ({1, −1}, ×) .
D2n /hai =
Example 17.3. Let R+ denote the set of positive real numbers. Then R+ is a group
under multiplication.
Let ϕ : (R \ {0}, ×) → (R+ , ×) be the map given by ϕ(x) = |x| for x ∈ R \ {0}. If
x, y ∈ R \ {0}, then |xy| = |x||y|, and so ϕ is a homomorphism. The kernel is {1, −1}
and the image is R+ , so {1, −1} E R \ {0} and
∼ R+ .
R \ {0}/{1, −1} =
Example 17.4. Let H be the cyclic subgroup of G = Z×Z generated by (2, 3). Since G is
abelian, every subgroup, and in particular H, is normal, so we can form the quotient
∼ Z.
group G/H. We’ll show that G/H =
This will follow from the Homomorphism Theorem if we can find a surjective
homomorphism ϕ : G → Z with ker(ϕ) = H.
Define ϕ : G → Z by ϕ(m, n) = 3m − 2n. Then for (m, n), (m 0 , n 0 ) ∈ G,
ϕ ((m, n) + (m 0 , n 0 )) = ϕ(m + m 0 , n + n 0 )
= 3(m + m 0 ) − 2(n + n 0 )
= ((3m − 2n) + (3m 0 − 2n 0 )
= ϕ(m, n) + ϕ(m 0 , n 0 )
so ϕ is a homomorphism (the group operation is addition for all the groups concerned).
60
COURSE NOTES 2016-2017
It’s easy to check that ϕ is surjective and that its kernel is H, so by the Homomorphism Theorem
∼ im(ϕ) = Z.
G/H = G/ ker(ϕ) =
We could replace (2, 3) with any element (a, b) where hcf(a, b) = 1.
Example 17.5. We’ll give a new proof that if hcf(m, n) = 1 then
∼ (Z/mZ) × (Z/nZ)
Z/mnZ =
(This follows from what we proved earlier about direct products of cyclic groups.)
Define
ϕ : Z → (Z/mZ) × (Z/nZ)
by ϕ(a) = ([a]m , [a]n ). Then for a, b ∈ Z,
ϕ(a + b) = ([a + b]m , [a + b]n ) = ([a]m , [a]n ) + ([b]m , [b]n ) = ϕ(a) + ϕ(b),
so ϕ is a homomorphism.
The kernel of ϕ is mnZ, since
a ∈ ker(ϕ) ⇔ ([a]m , [a]n ) = ([0]m , [0]n )
⇔ a is divisible by both m and n
⇔ a is divisible by mn,
since hcf(m, n) = 1.
Again because of the fact that hcf(m, n) = 1, ϕ is also surjective, and so the Homomorphism Theorem gives
∼ (Z/mZ) × (Z/nZ).
Z/mnZ =
INTRODUCTION TO GROUP THEORY (MATH 10005)
61
18. G ROUP ACTIONS
In this section we come full circle. We started by introducing the group of symmetries of an object as a way of understanding the object. But it turns out that one of the
most powerful ways to understand a group is to find different objects that it “acts” as
symmetries of in different ways. Here we’ll just cover the basic ideas, but this is one
of the main tools in more advanced group theory, such as you will see if you take the
third year Group Theory unit.
The idea of a group action generalizes the idea that the symmetric group Sn “acts”
on the set {1, . . . , n} by permuting the elements. More generally we can replace the
symmetric group be any group, and the set {1, . . . , n} by any set, and we are led to
the following definition.
Definition 18.1. A group action (or just action) of a group G on a set X is a function
G × X → X, where we’ll denote the image of (g, x) as g · x, such that
(1) e · x = x for all x ∈ X, and
(2) g · (h · x) = (gh) · x for all g, h ∈ G and x ∈ X.
Remark 18.1. The idea is that we think of the group element g as “acting” on the
element x ∈ X, and sending it to another element g · x ∈ X. Property (1) says that the
identity e ∈ G fixes every element of X, and property (2) says that “acting” on x by
h and then by g is the same as acting by the single element gh, thus connecting the
group structure of G with the way it can act on X.
Example 18.1. Sn acts on {1, . . . , n} by σ · x = σ(x).
Example 18.2. The dihedral group D2n acts on the set of vertices of a regular n-sided
polygon. For example, if we label the vertices anticlockwise 1, . . . , n and, as usual, a
is a rotation anticlockwise through an angle of 2π/n and b is a reflection in the line
of symmetry through vertex 1, then
•
•
•
•
•
•
a·1=2
a·2=3
a·n=1
b·1=1
b·2=n
b · n = 2.
Example 18.3. The dihedral group D8 acts on the set of two diagonals of a square. If
we denote by C the diagonal joining vertices 1 and 3, and by D the diagonal joining
vertices 2 and 4, then, for example,
• a·C=D
• a·D=C
62
COURSE NOTES 2016-2017
• b·C=C
• b·D=D
This example and the previous one show that the same group can have natural actions on more than one set.
The next two examples show how actions arise naturally even for groups that are
not necessarily defined as groups of symmetries.
Example 18.4. Let G be any group (with the group operation written multiplicatively),
then G acts on itself (i.e., we’re taking X = G) by
g · x = gx for g, x ∈ G.
Here the properties required for a group action follow from the definition of a group:
(1) e · x = ex = x by definition of the identity element, and
(2) g · (h · x) = g(hx) = (gh)x = (gh) · x by associativity.
Example 18.5. Let G be a group and H ≤ G a subgroup, and let X be the set of left
cosets of H in G. Then
g · (xH) = gxH
for g, x ∈ G defines an action of G on X, since
(1) e · (xH) = exH = xH for and xH ∈ X, and
(2) g · (h · (xH)) = g · (hxH) = ghxH = (gh) · (xH) for g, h ∈ G and xH ∈ X.
Lemma 18.1. If G acts on X, then, for each g ∈ G, the map X → X defined by x 7→ g · x is
bijective. In particlar, if x, y ∈ X and g · x = g · y for any g ∈ G then x = y.
Proof. For x ∈ X and g ∈ G,
x = e · x = (g−1 g) · x = g−1 · (g · x)
and
x = e · x = (gg−1 ) · x = g · (g−1 · x),
so the function x 7→ g−1 · x is inverse to the function x 7→ g · x.
Remark 18.2. In fact, the map from G to S(X) (the group of permutations of X) that
sends g to the permutation x 7→ g · x is a group homomorphism. We won’t pursue
this point of view here, but it is quite straightforward to prove.
There are natural and important subsets of G and X associated with an element of
X when G acts on X.
Definition 18.2. Suppose G acts on X and x ∈ X. Then
(1) the orbit G · x of x is
{g · x : g ∈ G} ⊆ X,
and
INTRODUCTION TO GROUP THEORY (MATH 10005)
63
(2) the stabilizer Gx of x is
{g ∈ G : g · x = x} ⊆ G.
Example 18.6. Let G = Sn act on X = {1, . . . , n} in the usual way. Then G · x = X
for every x ∈ X, since for every y ∈ X there is a permutation σ with σ(x) = y.
The stabilizer of x = n is the group of permutations that fix n, and just permute
{1, . . . , n − 1}.
Example 18.7. Let G = D8 act on the set X of vertices of a square, labelled 1, . . . , 4 as
usual. The orbit of every vertex is the whole of X, since for any two vertices, there is a
symmetry (in fact, a rotation), taking one to the other. There are two symmetries that
fix the vertex 1: the identity and the reflection b, so G1 = {e, b}. There are also two
symmetries fixing the vertex 2: the identity and the reflection in the diagonal joining
vertices 2 and 4, which is a2 b, so G2 = {e, a2 , b}.
Example 18.8. Taking the previous example, it’s clear that any subgroup of D8 also
acts on the set of vertices of the square. Let H = hbi act. Then the orbit of 1 is just
{1} and the orbit of 3 is just {3}, since every element of H fixes 1 and 3. But b swaps
vertices 2 and 4, so the orbits of 2 and 4 are both {2, 4}. H is the stabilizer of both 1
and 3, but the stabilizers of 2 and 4 are both the trivial group {e}.
Proposition 18.2. Let G act on X, and let x, y ∈ X. Then
G · x ∩ G · y 6= ∅ ⇔ G · x = G · y.
I.e., two orbits (for the same action) are either disjoint or equal.
Proof. Suppose G · x ∩ G · y 6= ∅. Then we can choose z ∈ G · x ∩ G · y, so that
z = g · x = h · y for some g, h ∈ G. Then
x = g−1 · (g · x) = g−1 · z = g−1 · (h · y) = (g−1 h) · y
and
y = h−1 · (h · y) = h−1 · z = h−1 · (g · x) = (h−1 g) · x.
If k ∈ G, then
k · x = k · (g−1 h) · y = (kg−1 h) · y ∈ G · y
and
k · y = k · (h−1 g) · x = (kh−1 g) · x ∈ G · x,
so G · x = G · y.
Conversely, if G · x = G · y then G · x ∩ G · y contains x = e · x, so is certainly not
empty.
64
COURSE NOTES 2016-2017
Remark 18.3. This means that the set X is partitioned into the orbits, so that each
element of X is in one and only one orbit. For example, for the action of hbi on the
set of vertices of a square, we saw that the orbits are
{1}, {2, 4} and {3}.
Theorem 18.3. Let G act on X, and let x ∈ X. Then the stabilizer Gx is a subgroup of G.
Proof. First, e · x = x by property (1) in the definition of a group action, so e ∈ Gx .
Suppose g, h ∈ Gx . Then by property (2) in the definition of a group action
(gh) · X = g · (h · x)
= g · x since h ∈ Gx
= x since g ∈ Gx ,
and so gh ∈ Gx , so Gx is closed under multiplication.
Finally, if g ∈ Gx then
g−1 · x = g−1 · (g · x) since x ∈ Gx
= (g−1 g) · x
=e·x
= x,
and so g−1 ∈ Gx . So Gx is closed under taking inverses.
So Gx ≤ G.
One of the most useful basic results about group actions is the following theorem,
that can be regarded as a generalization of Lagrange’s Theorem (we’ll explain how
this works after proving the theorem).
Theorem 18.4 (Orbit-Stabilizer Theorem). Let G be a group acting on a set X, and let
x ∈ X. Then
|G : Gx | = |G · x|,
so that, if G is finite.
|G| = |G · x||Gx |.
Proof. Recall that the index |G : Gx | is the number of left cosets of Gx in G, so we can
prove the first equation by finding a bijection between the set of left cosets and the
set G · x.
Define α : G → G · x by α(g) = g · x. Then, for g, h ∈ G,
α(g) = α(h) ⇔ g · x = h · x
⇔ (h−1 g) · x = x
⇔ h−1 g ∈ Gx
⇔ gGx = h(h−1 g)Gx = hGx ,
INTRODUCTION TO GROUP THEORY (MATH 10005)
65
so α induces a bijection gGx 7→ g · x between the set of left cosets and the set G · x.
The second equation follows since, if G is finite, |G| = |G : Gx ||Gx |.
Example 18.9. Let G = Sn act on {1, . . . , n} in the usual way. Then G · n = {1, . . . , n},
so |G · 1| = n, and Gn is the group of permutations of {1, . . . , n − 1}, which has order
(n − 1)!.
So
|G| = n! = n ((n − 1)!) = |G · n||Gn |.
Example 18.10. Let H ≤ G be finite groups and let G act on the set X of left cosets of
H in G as described above. Then H = eH is an element of X with stabilizer H (since
g · H = gH is equal to H if and only if g ∈ H, and with orbit X with |G : H| elements.
So the Orbit-Stabilizer Theorem just says that
|G| = |G : H||H|,
which is just Lagrange’s Theorem.
Example 18.11. Let G = D2n acting on the set of vertices of a regular n-sided polygon
in the usual way, whhere b is the reflection in the line of symmetry through vertex 1.
Then the orbit G · 1 is the set of all n vertices, and the stabilizer G1 is the subgroup
{e, b}. So the Orbit-Stabilizer Theorem says
|G| = 2n = |G · 1||G1 |.
Notice that if we didn’t already know the order of D2n , this would give a method of
calculating it, which is actually very useful in more complicated examples.
c
University
of Bristol 2017. This material is copyright of the University unless
explicitly stated otherwise. It is provided exclusively for educational purposes at the
University and is to be downloaded or copied for your private study only.