14 1. Sylow Theory

1C

We are now ready to study in greater detail the nonempty set Sylp(G) of

Sylow p-subgroups of a finite group G.

1.11. Theorem. Let P be an arbitrary p-subgroup of a finite group G, and

suppose that S G Sylp(G). Then

PCS9

for some element g G G.

Proof. Let O = {Sx \ x G G}, the set of right cosets of S in G, and note

that \Q\ — \G : S\ is not divisible by p since S is a Sylow ^-subgroup of G.

We know that G acts by right multiplication on fi, and thus P acts too, and

Q is partitioned into P-orbits. Also, since \Q\ is not divisible by p1 there

must exist some P-orbit O such that \0\ is not divisible by p.

By the fundamental counting principle, \0\ is the index in P of some

subgroup. It follows that \0\ divides |P|, which is a power of p. Then \0\ is

both a power of p and not divisible by p, and so the only possibility is that

\0\ = 1. Recalling that all members of ft are right cosets of S in G, we can

suppose that the unique member of O is the coset Sg.

Since Sg is alone in a P-orbit, it follows that it is fixed under the action

of P, and thus Sgu = Sg for all elements u G P. Then gu G Sg, and hence

u G s "

1

^ = S*. Thus P C S*, as required. •

If S is a Sylow ^-subgroup of G, and g G G is arbitrary, then the conju-

gate S^ is a subgroup having the same order as S. Since the only requirement

on a subgroup that is needed to qualify it for membership in the set Sylp(G)

is that it have the correct order, and since S G Sylp(G) and

\S9\

= |5|, it

follows that

Sg

also lies in Sylp(G). In fact every member of Sylp(G) arises

this way: as a conjugate of S. This is the essential content of the Sylow

conjugacy theorem. Putting it another way: the conjugation action of G on

Sylp(G) is transitive.

1.12. Theorem (Sylow C). If S andT Sylow p-subgroups of a finite group

G, then T —

S9

for some element g G G.

Proof. Applying Theorem 1.11 with T in place of P, we conclude that

T C

S9

for some element g G G. But since both S and T are Sylow p-

subgroups, we have \T\ — \S\ —

\Sg\,

and so the containment of the previous

sentence must actually be an equality. •

The Sylow C-theorem yields an alternative proof of Problem IB. 1(b),

which asserts that if a group G has a normal Sylow p-subgroup 5, then

S is the only Sylow p-subgroup of G. Indeed, by the Sylow C-theorem, if

T G Sylp(G), then we can write T —

S9

= 5, where the second equality is a

consequence of the normality of S.