16 1. Sylow Theory

Proof. Since np(G) = |Sylp(G)| is the total number of conjugates of S in

G, the result follows by Corollary 1.6. •

In particular, it follows that np(G) divides |G|, but we can say a bit

more. If S e Sylp(G), then of course, S C NG(S) since S is a subgroup, and

thus \G : S\ = \G : N

G

(5)||N

G

(5) : S\. Also, np(G) = \G : N

G

(5)|, and

hence np(G) divides \G : S\. In other words, if we write \G\ =

pam,

where

p does not divide m, we see that np(G) divides m. (We mention that the

integer m is often referred to as the ^/-part of |G|.)

The information that np(G) divides the ^/-part of \G\ becomes even more

useful when it is combined with the fact (probably known to most readers)

that np(G) = 1 mod p for all groups G. In fact, there is a useful stronger

congruence constraint, which may not be quite so well known. Before we

present our theorem, we mention that if 5, T G Sylp(G), then |5| = |T|, and

thus \S: SHT\ = \S\/\SnT\ = |T|/|SnT | = \T:SnT\. The statement of

the following result, therefore, is not really as asymmetric as it may appear.

1.16. Theorem. Suppose that G is a finite group such that np(G) 1; and

choose distinct Sylow p-subgroups S and T of G such that the order \S DT\

is as large as possible. Then np(G) = 1 mod \S : S H T\.

1.17. Corollary. If G is a finite group and p is a prime, then np{G) = 1

mod p.

Proof. If np(G) — 1, there is nothing to prove. Otherwise, Theorem 1.16

applies, and there exist distinct members S,T £ Sy\p(G) such that np(G) =

1 mod \S : SnT\, and thus it suffices to show that \S : SDT\ is divisible by

p. But \S : S fl T\ = \T : S D T\ is certainly a power of p, and it exceeds 1

since otherwise S = S HT = T, which is not the case because S and T are

distinct. •

In order to see how Theorem 1.16 can be used, consider a group G of

order 21,952 =

26»73.

We know that 77, 7 must divide

26

= 64, and it must

be congruent to 1 modulo 7. We see, therefore, that 717 must be one of 1,

8 or 64. Suppose that G does not have a normal Sylow 7-subgroup, so that

nj 1. Since neither 8 nor 64 is congruent to 1 modulo

72

= 49, we see

by Theorem 1.16 that there exist distinct Sylow 7-subgroups S and T of G

such that \S :SHT\ = 7.

Let's pursue this a bit further. Write D — S fl T in the above situation,

and note that since \S : D\ — 7 is the smallest prime divisor of |iS| =

73,

it follows by Problem 1A.1, that D S. Similar reasoning shows that also

D T, and hence S and T are both contained in T V =

NG(D). NOW

S

and T are distinct Sylow 7-subgroups of A/", and it follows that 727(N) 1,

and hence n7(N) 8 by Corollary 1.17. Since n7(N) is a power of 2 that