Page 136 - DMTH403_ABSTRACT_ALGEBRA

Unit 12: Sylow’s Theorems

12.1.1 A Proof of Sylow’s Theorems Notes

In this handout, we give proofs of the three Sylow theorems which are slightly different from
the ones in the book. Recall the following lemma:
Lemma: Let p be a prime number, and let G be a p-group (a finite group of order p for some
k
k  1) acting on a finite set S. Then the number of fixed points of the action is congruent to |S|
modulo p.

We make the following definition: if G has order p m with p | m, a Sylow p-subgroup of G is
k
a subgroup of order p .
k
Theorem (Sylow’s First Theorem): If G is a finite group of order n = p m with p prime and
k
p | m, then G has a subgroup of order p . In other words, if Syl (G) denotes the set of Sylow
k
p
p-subgroups of G, then Syl (G)  0 .
p
Proof. The proof is by induction on |G|, the base case |G| = 1 being trivial. If there exists a
proper subgroup H of G such that p | [G : H], then a Sylow p-subgroup of H is also a a Sylow
p-subgroup of G and we’re finished by induction. So without loss of generality, we may assume
that p | [G : H] whenever H < G. From the class equation, it follows that p | |Z |. By Cauchy’s
G
theorem, there exists a subgroup N  Z of order p, which is necessarily normal in G. Let G =
G
G/N, so | G | = p m. By induction, G has a subgroup P of order p . Let P be the subgroup
k–1
k–1
of G containing N which corresponds to P by the first isomorphism theorem. Then
k–1 .
.
|P| = |P/N| |N| = p p = p ,
k
so that P is a Sylow p-subgroup of G as desired.
Theorem (Sylow’s Second Theorem): If G is a finite group and p is a prime number, then all
Sylow p-subgroups of G are conjugate to one another.

Proof: We show more precisely that if H is any subgroup of G of p-power order and P is any
Sylow p-subgroup of G, then there exists x  G such that H  xPx . (This implies the theorem,
–1
since if H  Syl (G) then |H| = |P| = |xPx |, which implies that H = xPx , so that H is conjugate
–1
–1
p
to P.) Note that H acts on G/P (the set of left cosets of P in G) by left multiplication. Let Fix denote
the elements of G/P fixed by this action. Then |Fix|  |G/P| (mod p) by the Lemma. Since
p | m = |G/P|, |Fix|  0, and thus Fix  0 ;. Let xP be a left coset fixed by the action. Then
hxP = xP  h  H  x Hx  P,
–1
so that H  xPx as desired.
–1
Theorem (Sylow’s Third Theorem): If G is a finite group and p is a prime number, let
n = |Syl (G)|. Then n | |G| and n  1 (mod p).
p
p
p
p
Proof: We consider the action of G on Sylp(G) by conjugation. By the second Sylow theorem, this
action is transitive, so there is just one orbit. Hence n , which is the size of this orbit, divides |G|.
p
To prove the congruence n  1 (mod p), we fix a Sylow p-subgroup P  Syl (G) and consider the
p
p
action of P on Syl (G) by conjugation. Let Fix denote the set of fixed points of this action. Note
p
that Q  Fix  P  N (Q), and in particular P  Fix. If Q  Fix, then P, Q  N (Q) are both Sylow
G
G
p-subgroups of N (Q), so they are conjugate in N (Q) (again by the second Sylow theorem). But
G
G
Q is a normal subgroup of N (Q), so P = Q. Thus Fix = {P}, and in particular |Fix| = 1. By the
G
Lemma, np  1 (mod p) as desired.
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