14 1. Sylow Theory
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
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 "
^ = 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
= |5|, it
follows that
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
for some element g G G.
Proof. Applying Theorem 1.11 with T in place of P, we conclude that
for some element g G G. But since both S and T are Sylow p-
subgroups, we have \T\ \S\
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
= 5, where the second equality is a
consequence of the normality of S.
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