Download §2 Group Actions Definition. Let G be a group, and Ω a set. A (left

§ 2 Group Actions
Definition. Let G be a group, and Ω a set. A (left) action of G on Ω is a map ψ : G × Ω −→ Ω, such that
1. ψ(e, x) = x for all x ∈ Ω, and
2. ψ(g, ψ(h, x)) = ψ(gh, x) for all g, h ∈ G and x ∈ Ω.
We shall generally write gx for ψ(g, x), except where this leads to ambiguities, or where other notation
is more convenient. By the second axiom, we may unambiguously write ghx without bracketing.
[A right action of G on Ω is defined similarly, but with the group elements written on the right instead of
the left; this is not purely a notational difference, as the second axiom in this case becomes ψ(ψ(x, g), h) =
ψ(x, gh). It can easily be checked that if x 7→ gx is a left action, then x 7→ xg−1 defines a right action, and
vice versa.]
Examples.
1.
The dihedral group D2n acts on a regular n-gon. (Take Ω to be the set of vertices, for instance.)
2.
The symmetric group Sn acts on the set {1, . . . , n}.
3.
The general linear group GLn (R) acts on Rn , considered as column vectors. (It also has a right action
on row vectors.)
Definition. Let G be a group acting on a set Ω. Let x ∈ Ω.
1.
The orbit of x, written OrbG (x), is the set {gx | g ∈ G}.
2.
The stabilizer of x, written StabG (x), is the set {g ∈ G | gx = x}.
Note that OrbG (x) is a subset of Ω, while StabG (x) is a subset of G.
Proposition 9. Let G be a group acting on a set Ω, and let x ∈ Ω. Then StabG (x) ≤ G.
Proof. Suppose that g, h ∈ StabG (x). Then ghx = g(hx) = gx = x, and so gh ∈ StabG (x). Also g−1 x =
g−1 (gx) = ex = x, and so g−1 ∈ StabG (x). So StabG (x) is a subgroup of G.
Theorem 10. z Orbit-Stabilizer Theorem
Let G be a finite group acting on a set Ω, and let x ∈ Ω. Then
|G| = | OrbG (x)|| StabG (x)|.
Proof. Let S = StabG (x). Let g, h ∈ G. We observe that
gx = hx ⇐⇒ h−1 gx = x ⇐⇒ h−1 g ∈ S ⇐⇒ gS = hS.
It follows that the number of distinct elements in OrbG (x) is equal to the number of cosets gS for g ∈ G. But
this is |G|/|S|, and the theorem follows.
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Proposition 11. Let G be a group acting on a set Ω. Then the relation R on Ω defined by R(x, y) if and only
if y ∈ OrbG (x) is an equivalence relation, whose equivalence classes are the orbits of G on Ω. (Briefly: the
orbits of G partition Ω.)
Proof. We need to show that R is reflexive, symmetric, and transitive.
Reflexive. We have ex = x for all x ∈ Ω, and so R(x, x).
Symmetric. Suppose R(x, y). Then y = gx for some g ∈ G. Now g−1y = g−1 gx = x, and so R(y, x).
Transitive. Suppose R(x, y) and R(y, z). Then there are g, h ∈ G such that y = gx and z = hy. But now
(hg)x = h(gx) = h = z, and so R(x, z).
Recall that for a set Ω, the group Sym(Ω) consists of all permutations on Ω (i.e. invertible functions
Ω −→ Ω) under composition.
Proposition 12. Let G be a group acting on a set Ω. Then for all g ∈ G the map ϕg : x 7→ gx is a permutation
of Ω. Moreover, the map ϕ : g 7→ ϕg is a homomorphism G −→ Sym(Ω).
Certainly ϕg is a function Ω −→ Ω, and it is a permutation since it has the inverse ϕg−1 . We note that
ϕgh (x) = (gh)x = g(hx) = ϕg ◦ ϕh (x), and so ϕ is a homomorphism as required.
Definition. Let G be a group acting on a set Ω.
1.
We say that the action of G is transitive if OrbG (x) = Ω for any x ∈ Ω.
2.
The kernel of the action is the kernel of the homomorphism ϕ from Proposition 12; that is to say, the
set {g ∈ G | gx = x for all x ∈ Ω}.
3.
We say that the action is faithful if its kernel is {e}.
A. Four important group actions
In this section, let G be any group. We introduce four important actions of G. Every one of these actions
plays an important part in the general theory of groups.
Action 1: action of G on itself by (left) translation.
This is an action of G on itself, i.e. we have Ω = G; it is sometimes called the left regular action. It is given
by (g, x) 7→ gx, where gx here denotes multiplication in G. So for this action, our usual notation for a group
action is in happy agreement with our standard notation for group multiplication. It is easy to show that this
is indeed an action, since ex = x for all x ∈ G, and since (gh)x = g(hx) by the associativity axiom for groups.
There is a similar right action given by (x, g) = xg; this is the right regular action.
Let x, y ∈ G. If g = yx−1 , then y = gx. Hence OrbG (x) = G, and so this action is transitive. The action is
faithful, since clearly its kernel is {e}; in fact StabG (x) = {e} for every x ∈ G. This gives an easy proof of
the following theorem.
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Theorem 13. z Cayley’s Theorem
Let G be a finite group. Then G is isomorphic to a subgroup of Sn for some n.
Proof. By Proposition 12 there is a homomorphism ϕ : G −→ Sym(G) whose kernel is {e}, the kernel of the
action defined above. Now G ∼
= G/{e} ∼
= Im ϕ by the First Isomorphism Theorem. But Im ϕ is a subgroup
of Sym(G), and Sym(G) ∼
= Sn , where n = |G|.
Example. Cayley’s Theorem predicts that the action of S3 on itself by translation should give an injective
homomorphism of S3 into S6 (since |S3 | = 6). To avoid a notational clash, we let S3 act on {a, b, c} instead
of the usual {1, 2, 3}. We shall assign each element of S3 a number, 1, . . . , 6, as follows:
e = 1;
(a b) = 2;
(a c) = 3;
(b c) = 4;
(a b c) = 5;
(a c b) = 6.
The map x 7→ ex clearly leaves each x fixed, and so the permutation effected on {1, . . . , 6} is the identity.
Next consider the map x 7→ (a b)x; we calculate
(a b)1 = 2;
(a b)2 = 1;
(a b)3 = 6;
(a b)4 = 5;
(a b)5 = 4;
(a b)6 = 3.
So translation by (a b) effects the permutation (1 2)(3 6)(4 5) on the (numbered) elements of S3 . The remaining elements of S3 can be treated similarly; we obtain the homomorphism
eS3
7→ eS6 ,
(a b) 7→ (1 2)(3 6)(4 5),
(b c) 7→ (1 4)(2 5)(3 6),
(a b c) 7→ (1 5 6)(2 3 4),
(a c) 7→ (1 3)(2 5)(4 6),
(a c b) 7→ (1 6 5)(2 4 3).
(It should be pointed out that this is far being from the only injective homomorphism of S3 into S6 .)
Action 2: action of G on the (left) cosets of a subgroup.
Let H be a subgroup of G. For this action we take Ω = {xH | x ∈ G}; the action is given by (g, xH) 7→ gxH.
It is easy to show that this is indeed an action. (There is a similar right action of G on the right cosets, given
by (Hx, g) = Hxg.)
Let xH, yH ∈ Ω. Then putting g = yx−1 , we have gxH = yH. Hence yH ∈ OrbG (xH) for all x, y, and so
the action is transitive.
For any xH ∈ Ω, we have
gxH = xH ⇐⇒ gx ∈ xH ⇐⇒ g ∈ xHx−1 .
So we have StabG (xH) = xHx−1 .
Action 3: action of G on itself by conjugation.
This is another action for which Ω = G, but it is very different from the left regular action. It is given by
(g, x) 7→ gxg−1 . This is an action for which we cannot use our standard notation gx for the action of g on
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x, since this would conflict with group multiplication. Instead we write g x for gxg−1 , which we call the
conjugate of x by g.
This defines an action, since e x = exe−1 = x, and
gh
x = (gh)x(gh)−1 = ghxh−1 g−1 = g (hxh−1 = g (h x).
−1
There is a similar right action given by (x, g) 7→ g−1 xg. We may write xg for g−1 xg; note that xg = g x.
The orbits of the conjugacy action are known as conjugacy classes. We write G x, or else ConG (x), for
the conjugacy class of x. The stabilizer of x is the subgroup {g ∈ G | gx = xg}. This is known as the
centralizer of x in G, and written CentG (x). By the Orbit-Stabilizer Theorem, if G is finite, then |G| =
| ConG (x)|| CentG (x)|.
This action is never transitive if |G| > 1, since the identity e lies in a conjugacy class on its own. The
kernel of the action is the subgroup {g ∈ G | gx = xg for all x ∈ G}. This is known as the centre of G, and
written Z(G).
Action 4: action of G on its subgroups by conjugation.
Let Ω be the set of all subgroups of G. Then G acts on Ω by (g, H) 7→ gHg−1 . We have seen that gHg−1 is
the stabilizer of gH under Action 2; so it is a subgroup of G, called. We write gH for gHg−1 , and call it the
conjugate of H by g. It is easy to check that the map is an action of G.
A subgroup H of G is normal if and only if OrbG (H) = {H}. The stabilizer of a subgroup H is known
as the normalizer of H in G, and written NG (H). It is the largest subgroup of G of which H is a normal
subgroup.
§ 3 Sylow’s Theorems
Let G be a group with finite order n. Lagrange’s Theorem tells us that the order of any subgroup of G
is a divisor of n. It is not in general the case that G has a subgroup of order d for every divisor d of n. For
instance, the group A4 , which has order 12, has no subgroup of order 6. In this section, however, we show
that in the special case that d is a power of a prime number, G does have a subgroup of order d.
Definition. Let p be a prime number.
1.
A finite group whose order is a power of p is said to be a p-group.
2.
If G is a group, and H a subgroup of G which is a p-group, then H is a p-subgroup of G.
3.
An element of G whose order is a power of p is a p-element of G.
We deal first with subgroups of prime order.
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Theorem 14. z Cauchy’s Theorem
Let p be a prime, and G a finite group whose order is divisible by p. Then G has an element of order p.
Note that if g has order p, then hgi is a subgroup of order p. So an equivalent statement of this theorem
is that a group whose order is divisible by p has a subgroup of order p.
Proof. Let Ω be the set of p-tuples of elements of G whose product is the identity; i.e.
Ω = {(g1 , . . . , g p ) ∈ G p | g1 . . . g p = e}.
Notice that the tuple (g1 , . . . , g p ) belongs to Ω if and only if g p = (g1 . . . g p−1 )−1 . This observation has two
important consequences. Firstly, it allows us to calculate the size of Ω as |G| p−1 , since for any choice of
g1 , . . . , g p−1 there is a unique g p such that (g1 , . . . , g p ) ∈ Ω. Secondly, it shows that
(g1 , . . . , g p ) ∈ Ω ⇐⇒ (g p , g1 , . . . , g p−1 ) ∈ Ω,
since inverses in groups are two-sided. Let θ be a generator of the cyclic group Cp . Then there is an action
of Cp on Ω given by θ (g1 , . . . , g p ) = (g p , g1 , . . . , g p−1 ). By the Orbit-Stabilizer Theorem, every orbit of this
action has size divisible by p. If there are M orbits of size 1, and N orbits of size p, then we clearly have
|Ω| = M + pN. But |Ω| = |G| p−1 is divisible by p (since |G| is), and it follows that p divides M. Clearly the
orbit of (e, . . . , e) has size 1, and so M > 0. Therefore there exists at least one other orbit of size 1. But the
unique element of this orbit is fixed by θ , and so it has the form (g, . . . , g) for some g ∈ G such that g p = e.
Now g is an element of order p in G, as required.
The principal theorems of this chapter, named collectively after the Norwegian mathematician Ludwig
Sylow, deal with the case of subgroups whose order is the largest power of a prime p which divides |G|.
These are some of the most important theorems in group theory. As with the Isomorphism Theorems in
Section 1, there is no standard way of numbering these results; some authors choose to collect them together
into a single theorem.
Theorem 15. z First Sylow Theorem
Let G be a finite group of order n, let p be prime, and let pa be the largest power of p dividing n. Then G
has a subgroup of order pa .
Definition. Let G be a finite group. If p is prime, and pa the largest power of p dividing |G|, then a subgroup
of G whose order is pa is known as a Sylow p-subgroup of G. We write Syl p (G) for the set of Sylow psubgroups of G.
Theorem 16. z Second Sylow Theorem
Let G be a finite group, and p a prime. Then | Syl p (G)| ≡ 1 mod p.
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Remark. It is clear that Theorem 16 implies Theorem 15, since if | Syl p (G)| is congruent to 1 mod p then
it cannot be 0. However the logical structure of our argument will be to establish Theorem 15 first, and to
derive Theorem 16 and the other Sylow Theorems as consequences.
Theorem 17. z Third Sylow Theorem
Let G be a finite group, and p a prime. Every p-subgroup of G is contained in at least one Sylow p-subgroup.
Theorem 18. z Fourth Sylow Theorem
Let G be a finite group, and p a prime. The Sylow p-subgroups of G form a single conjugacy class of
subgroups, i.e. if P, Q ∈ Syl p (G) then there exists g ∈ G such that Q = g P.
Before embarking on the proof of the First Sylow Theorem, we recall that Z(G), the centre of G, is
defined to be {g ∈ G | gh = hg for all h ∈ g}. It is clear that an element g of G lies in a conjugacy class
of size 1 if and only if g ∈ Z(G), since for any h ∈ G we have h g = g ⇐⇒ gh = hg. For this reason, the
conjugacy classes of size 1 are known as the central conjugacy classes.
If G is a finite group, then it has finitely many conjugacy classes, and so finitely many non-central
classes. Suppose that C1 , . . .Ck are the non-central conjugacy classes of G; so |Ci | > 1 for all i. Then since G
is partitioned by its conjugacy classes, and since the central conjugacy classes are those which lie in Z(G),
we have the Class Equation for G:
k
|G| = |Z(G)| +
∑ |Ci |.
i=1
We are now in a position to begin the proof.
Proof of First Sylow Theorem. We proceed by induction on n, the order of G. Our inductive hypothesis is
that every group whose order is less than n possesses a Sylow p-subgroup. We show that a group of order n
has a Sylow p-subgroup.
If p does not divide n, then {e} is a Sylow p-subgroup (of order p0 ) for G. So we may suppose that the
highest power of p dividing n is pa where a > 0. We consider two cases:
Case i. p divides |Z(G)|. Then by Cauchy’s Theorem, Z(G) has a subgroup K of order p. Now observe that
if k ∈ K then gk = kg for all g ∈ G (since k is central), and so gK = Kg for all g. So K E G. The quotient
G/K has order n/p, which is less than n, and so by the inductive hypothesis, G/K has a Sylow p-subgroup
Q of order pa−1 . By Propositions 2 and 3 we see that there is a subgroup of P of G which contains K such
that P/K ∼
= Q. Now |P| = |K||Q| = pa , and so P is a Sylow p-subgroup of G.
Case ii. p does not divide |Z(G)|. Then consider the class equation: by assumption p divides |G|, but it
does not divide |Z(G)|. Therefore p does not divide ∑i |Ci |, and so there is at least one non-central conjugacy
class Ci whose size is not divisible by p. Let x ∈ Ci . We have |Ci || CentG (x)| = |G| by the Orbit-Stabilizer
Theorem, and it follows that pa divides | CentG (x)|. But since |Ci | > 1, we see that | CentG (x)| < n, and so by
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the inductive hypothesis, CentG (x) has a Sylow p-subgroup P. But |P| = pa , and so P is a Sylow p-subgroup
of G.
For the proofs of the other Sylow Theorems, the following proposition will be needed.
Proposition 19. Let G be a finite group, and let p be prime. Let P be a Sylow p-subgroup of G, and Q any
p-subgroup. Then either Q ≤ P, or else there exists q ∈ Q such that qPq−1 6= P.
An equivalent statement is: if P ∈ Syl p (G), and if Q is a p-subgroup of NG (P), then Q ≤ P.
Proof. Suppose qPq−1 = P for all q ∈ Q. Then qP = Pq for all q, and so QP = PQ. Hence PQ is a
subgroup of G of order |P||Q|/|P ∩ Q|, by Proposition 5. Clearly |P||Q| is a power of p; but since P is a
Sylow p-subgroup of G, the order of PQ cannot be greater than |P|. Hence |P ∩ Q| = |Q|, and so Q ≤ P as
required.
Proof of Second, Third and Fourth Sylow Theorems. Let G be a finite group, and p a prime. By the First
Sylow Theorem, G has a Sylow p-subgroup P. Let Ω be the set of subgroups of G which are conjugate to P
in G. Then every element of Ω is a Sylow p-subgroup of G.
Consider the action of P on Ω by conjugation. Clearly OrbP (P) has size 1. If P′ is another element of
Ω then P is not a subgroup of P′ , and so by Proposition 19 it does not normalize P′ . But since | OrbP (P′ )|
divides |P|, it is a power of p greater than 1, and so it is divisible by p. Therefore Ω is the union of P-orbits
whose sizes are divisible by p, together with one part of size 1, and it follows that |Ω| ≡ 1 mod p.
Let Q be a p-subgroup of G, and consider the action of Q on Ω. Since the size of any Q-orbit is a power
of p, and since the union of the orbits is Ω, whose size is 1 mod p, we see that there must be an orbit of size
1. So there exists some P′ ∈ Ω such that Q ∈ NG (P′ ). Now by Proposition 19 it follows that Q ≤ P′ . We
have shown that every p-subgroup lies in one of the Sylow p-subgroups in the set Ω, and this establishes the
Third Sylow Theorem. Furthermore, if the subgroup Q is a Sylow p-subgroup, then clearly we must have
Q = P′ ∈ Ω, and so Ω = Syl p (G). It follows that Syl p (G) is a single conjugacy class of subgroups of G,
whose size is 1 mod p, and this establishes the Second and Fourth Sylow Theorems.
We note that the Fourth Sylow Theorem has the following corollary.
Corollary 20. Let G be a finite group and p a prime. Then | Syl p (G)| divides |G|.
Proof. The Sylow p-subgroups form a single orbit under the conjugacy action of G; the size of an orbit
divides |G| by the Orbit-Stabilizer Theorem.
It remains to deal with p-subgroups which are not Sylow subgroups. We shall need the following fact.
Proposition 21. Let p be a prime, and let G be a non-trivial p-group. Then the center Z(G) is non-trivial.
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Proof. The size of any conjugacy class of G divides |G|, and is therefore a power of p. Consider the class
equation for G (given before the proof of the First Sylow Theorem). Since p divides |G|, and also |Ci | for
each i, we see that p divides |Z(G)|. This implies that |Z(G)| > 1.
Proposition 22. Let G be a finite group, and p a prime, and let pb be a power of p dividing |G|. Then G has
a subgroup of order pb .
Proof. Since any group has a Sylow p-subgroup, it will be sufficient to prove the theorem in the case that G
is a p-group. So let us suppose that |G| = pa . We proceed by induction on a; our inductive property P(a) is
that any group of order pa has a subgroup of order pb for all b < a. We observe that P(1) is certainly true;
this is the base case for the induction. Suppose as our inductive hypothesis that P(a) holds, and let G be a
group of order pa+1 . Let b < a + 1. If b = 0 then {e} is a subgroup of G of order pb , so we may assume that
b > 0.
By Proposition 21, the centre Z(G) is a non-trivial p-group. We now argue as in Case i. of the proof
of the First Sylow Theorem. By Cauchy’s Theorem, Z(G) has a subgroup K of order p, and K E G. The
quotient G/K has order pa , and so by the inductive hypothesis, G/K has a subgroup Q of order pb−1 . Now
by Propositions 2 and 3, there is a subgroup of P of G containing K, such that P/K ∼
= Q, and clearly P has
order pb as required.
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