Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

y conjugation). If there is a nontrivial class Cx such that |ZG(x)| is divisible by p k , then p k
is also the largest power of p that can possibly divide |ZG(x)| since ZG(x) is a subgroup in G.
Clearly |ZG(x)| < |G| because Cx is nontrivial, so by the inductive hypothesis this subgroup
already contains a Sylow p-group for G.
If no such x exists, then for every nontrivial class we have
#(elements in the class) = |G|
|ZG(x)|
is divisible by p .
because the multiplicity of p in the centralizer is less than that in G. Therefore by the class
equation, the number of elements in the center Z(G) must also be divisible by p and hence
|Z(G)| must include a factor of the form p s for some 1 ≤ s ≤ k. If s = k we simply apply the
abelian result to Z(G) and are done. Otherwise, consider the quotient group G/Z(G)p where
Z(G)p is the Sylow p-subgroup of the center, with |Z(G)p| = p s . (Any subgroup of the center
Z(G) is normal in G, so the quotient is a group.) Induction applies to this quotient, whose
order involves the prime factor p k−s . Therefore the quotient contains a Sylow p-subgroup Sp
of this order. The pullback Sp in G has order p k since we factor out a group of order p s to get
Spi. This Sp is the desired Sylow p-subgroup in G.
To prove (b) and (c) we use the following instructive lemma about actions of p-groups.
6.3.3 Lemma. Let X be a finite set acted on by p-group G. Let
X G = {x ∈ X : g · x = x, all g ∈ G}
be the set of G-fixed points in X. Then |X G | ≡ |X| (mod p).
Proof: The set X G is obviously G-invariant and so is the difference set X ∼ X G , which must
then be a union of disjoint nontrivial G-orbits. The cardinality |G|/|StabG(x)| of any such orbit
is a divisor of |G| and so must be a power p s with s ≥ 1. Hence |O| is congruent to 0 (mod p)
for every nontrivial orbit O in X. But then the same must be true for the union X ∼ X G of
these orbits. That means
as claimed. �
|X| = |X ∼ X G | + |X G | ≡ |X G | (mod p)
For (b) let n = p k m with gcd(m, p k ) = 1, fix a Sylow p-group Sp, and let H be a subgroup
whose order is a power of p. Consider the permutation action of H on the coset space G/Sp.
Then |Sp| = p k and |G/Sp| ≡/ 0 (mod p), so by 6.3.3 there is at least one fixed point xSp in X.
But then HxSp = xSp ⇒ x −1 HxSp = Sp ⇒ x −1 Hx ⊆ Sp and we’re done.
For (c) we fix a Sylow p-subgroup Sp and look at the action of Sp by conjugation on the
set X of all Sylow p-groups. The base point Sp is a fixed point in X; we claim that there
are no others, and then (c) follows from our lemma. If there were another fixed point H it
would be normalized by Sp in the sense that gHg −1 = H for g ∈ Sp. Thus Sp is contained in
the normalizer of H: the subgroup N = {g ∈ G : gHg −1 = H}. Hence Sp and H are Sylow
p-subgroups of N, and by (b) there exists some n ∈ N such that Sp = nHn −1 . By definition
of N this makes Sp = H as claimed.
Let Xp be the set of all p-Sylow subgroups and let P be any base point in Xp. By (b) the
action G × Xp → Xp is transitive so |Xp| = |G|/|StabG(P)| and |G| is a multiple of |Xp|, as
claimed. �
In 4.3.5 (Cauchy theorem) we proved that if p is a divisor of n = |G| then there are elements
in G such that o(x) = p. If p k is the largest power dividing n, one can generalize the Sylow
theorems to prove that there exist subgroups of order p r for every 1 ≤ r ≤ k.
Note: In analyzing the structure of groups, special interest attaches to the p-groups, for which
|G| = p k , so it is worth noting that
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