Sylow's First Theorem
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Ludwig
Sylow (1832-1918) was a Norwegian mathematician who discovered the first
deep theorems connecting number theory to the structure of finite groups.
He was not aware of group actions so his proofs were long and complicated
inductions, but nowadays these proofs are a breeze.
First we need a preliminary result.
Definition A p-group is a
group G whose order is a power of a prime p. It follows from Lagrange's
Theorem that if G is a p-group, then every element has p-power order and
from Cauchy's Theorem that if G is a group in which every element has p-power
order, then G is a p-group.
Notation Let G be a group
acting on a set X. Extending our previous notation, let Y be a subset
of X and H a subset of G. Define
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GY := {g in G: yg
= y for all y in Y}, the stabilizer of Y in G, a subgroup of G.
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XH := {x in X: xg
= x for all g in H}, the fixed set of H.
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YH := { yg : y in Y and g in
H}, so in particular, xG is just the orbit of x.
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In case H is a subgroup and Y is a union of orbits
of H, HY is the group H acting on the set Y.
Cauchy's Theorem is a first step towards a converse
of Lagrange's Theorem. A full converse would state that if n is a positive
integer dividing the order of a finite group G, then G has a subgroup of
order n. Of course this is false - we know that A5
has no subgroup of order 30, even though 30|60, because A5
is simple. Sylow's Theorem says that nevertheless the statement is true
if n is a prime power.
Sylow's First Theorem
Let G be a finite group of order prm with
r > = 1 and (p,m) = 1. Then G has a subgroup P of order pr.
Proof Let X = {E: E is a subset
of G with cardinality pr}. Define an action of G on X by Eg
= Eg := {eg: e in E}. [CHECK: this is a group action.]
Now |X| = {prm choose pr}
which is prime to p, since it is a product of pr factors of
the form (prm - k)/(pr - k), k = 0,1, ...,pr-1.
In each such factor, pn divides the numerator if and only if
pn divides the denominator. Since X is partitioned into orbits
of G, the length of at least one orbit, say Y, is not divisible by p. Choose
E in Y, and let P = GE = {g in G: Eg = E}. We certainly have
a subgroup of G. We must show its order is pr.
Since |Y| = |G|/|P|, we know pr||P|.
Now P is a group acting on E by right multiplication, so for all e in E,
Pe = {1}. Hence if Z is an orbit of P in E, |Z| = |P|, so |P|<
or = pr. Hence P has order pr.
The next part of Sylow's Theorem implies
that this P is conjugate in G to any other subgroup of the same order.
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Sylow's Second Theorem
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Let
G be a finite group of order prm with r > or = 1 and (p,m) =
1. Let P be a group constructed in Sylow's First Theorem. If H is any p-subgroup
of G, then H is a subgroup of g-1Pg for some g in G.
Proof As in the previous proof, let
E be an element of X contained in an orbit Y whose length is not divisible
by p. Consider the group H acting on Y. [CHECK: it does!] Since H is a
p-group, the length of each orbit of this action is a p-power, so there
are orbits of length 1. Hence for some F in Y, HF = H, so H
is a subgroup of GF. But F = Eg for some g in G, so GF
= g-1GEg, i.e. H is a subgroup of g-1Pg.
Corollary All groups of order pr
in G are conjugate.
Proof In Sylow's Second Theorem,
take H to be any subgroup of order pr. Then H is a subgroup
of g-1Pg of the same size, so they are equal.
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Sylow's Third Theorem
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A subgroup
of a group G of maximal p-power order is called a Sylow p-subgroup.
The next result tells us how many Sylow p-subgroups there are.
Sylow's Third Theorem
Let G be a finite group of order prm,
r >= 1, (p,m) = 1. Let n be the number of Sylow p-subgroups of G. Then
n|m and p|n-1.
Proof Let {Xi, i = 1,...,k}
be the set of G orbits in X (defined in the proof of Sylow's First Theorem).
If E in Xi and g in E, then 1 is an element of Eg-1
in Xi, i.e. each Xi contains a set E' containing
1. For each i = 1,2,...,k, choose Ei in Xi such that
1 in Ei and let Pi := G Xi.
For any g in Pi, g = 1g in (Ei)g
= Ei, so Pi is contained in Ei. On the
other hand, if x in Ei, then for all g in Pi, xg
in Ei, so x Pi is contained in Ei. Thus
Ei is a union of left cosets of Pi. Hence | Pi|
divides | Ei| = pr.
Thus if Pi is a Sylow p-subgroup
of G, then Pi = Ei and |X| = [G:Pi] =
m, while if not, then p|[G:Pi] = |Ei|.
Since G has n Sylow p-subgroups, there n
orbits Ei of length m, (one containing each Sylow p-subgroup),
and the remaining orbits have length divisible by p. Thus for some integer
j, {prm choose pr} = |X| = nm + jp, so for every
group G of order prm, {prm choose pr}
is congruent to nm modulo p. (Note that in this congruence, only n depends
on the group.)
For example, the cyclic group Z(prm)
has only one Sylow p-subgroup so {prm choose pr}
is congruent to m modulo p. Thus for any group G of order prm,
nm is congruent to m modulo p, and since (m,p) = 1, this implies n is congruent
to 1 modulo p.
Finally, to see why n|m, note that all Sylow
p-subgroups are conjugate, so n is just the number of conjugates of P,
so firstly, n|prm, and secondly, n = |[G:NG(P)]|.
But this number is prime to p, otherwise by Cauchy's Theorem G/NG(P)
would contain an element of order p, so P would not be a maximal p-subgroup
of G. Hence n|m.
The usefulness of Sylow's Theorems is due
to the fact that we know quite a lot about finite p-groups, for example:
Theorem 11.1 Let P be a finite p-group,
and let Q be a proper normal subgroup. Then Q 0
Z(P) (the centre of P) is non-trivial.
Proof Since Q is normal in P, P acts
on Q by conjugation. The orbits of length 1 form Q 0
Z(P), and there is at least one, namely {1}. But the sum of the lengths
of all the orbits is a p-power, so there must be at least p of length 1.
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