IB 9

not divisible by p. An alternative formulation of this definition relies on the

observation that every positive integer can be (uniquely) factored as a power

of the given prime p times some integer not divisible by p. In particular,

if we write \G\ =

pam,

where a 0 and p does not divide m 1, then a

subgroup S of G is a Sylow p-subgroup of G precisely when \S\ =

pa.

In

other words, a Sylow p-subgroup of G is a p-subgroup S whose order is as

large as is permitted by Lagrange's theorem, which requires that \S\ must

divide \G\. We mention two trivial cases: if \G\ is not divisible by p, then

the identity subgroup is a Sylow p-subgroup of G, and if G is a p-group,

then G is a Sylow p-subgroup of itself. The Sylow existence theorem asserts

that Sylow subgroups always exist.

1.7. Theorem (Sylow E). Let G be a finite group, and let p be a prime.

Then G has a Sylow p-subgroup.

The Sylow E-theorem can be viewed as a partial converse of Lagrange's

theorem. Lagrange asserts that if if is a subgroup of G and |if | = fc, then

k divides \G\. The converse, which in general is false, would say that if k

is a positive integer that divides |G|, then G has a subgroup of order k.

(The smallest example of the failure of this assertion is to take G to be the

alternating group A^ of order 12; this group has no subgroup of order 6.)

But if k is a power of a prime, we shall see that G actually does have a

subgroup of order k. If fc is the largest power of ,p that divides |G|, the

desired subgroup of order k is a Sylow p-subgroup; for smaller powers of p,

we will prove that a Sylow p-subgroup of G necessarily has a subgroup of

order k.

We are ready now to begin work toward the proof of the Sylow E-

theorem. We start with a purely arithmetic fact about binomial coefficients.

1.8. Lemma. Letp be a prime number, and leta0 andm 1 be integers.

Then

Proof. Consider the polynomial (1 +

X)p.

Since p is prime, it is easy to

see that the binomial coefficients (?) are divisible by p for 1 i p — 1,

and thus we can write (1 +

X)p

= 1 +

Xp

mod p. (The assertion that

these polynomials are congruent modulo p means that the coefficients of

corresponding powers of X are congruent modulo p.) Applying this fact a

2 2

second time, we see that (1 + X)

P

=

(1+XP)P

= l +

Xp

mod p. Continuing

like this, we deduce that (1 +

X)pa

= 1 +

Xpa

mod p, and thus

(1 +

X)payn

= (1 +

Xpa)m

mod p .