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\ =
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\ =
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.
Proof. Consider the polynomial (1 +
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 +
= 1 +
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)
= l +
mod p. Continuing
like this, we deduce that (1 +
= 1 +
mod p, and thus
(1 +
= (1 +
mod p .
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