Sylow Theorems
Sylow Theorems
• History
Peter Ludwig Mejdell Sylow (12 December 1832 –7 September
1918) was a Norwegian mathematician, who proved foundational
results in group theory. He was born and died in Christiania (now
Oslo). He 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.
He was a school teacher in Frederikshald, from 1858 to 1898. He
was a substitute lecturer at Christiania University in 1862, covering
Galois theory; he posed then the question that was to lead to the
Sylow subgroups and his theorems about them. He published the
Sylow theorems in 1872.
Subsequently he was an editor of Abel’s papers, with Sophus Lie.
He was appointed professor in 1898.
Sylow Theorems
• Introduction
Let G be a finite group of order n, say. Lagrange’s theorem says
that if H is a subgroup of G and |H| = d, then d is a divisor of n. It is
natural to ask whether the converse is true: given a divisor d of n,
can we find a subgroup H G such that |H| = d? The answer is no in
general; for example one can check that the group A4 has order 12
but there is no subgroup of order 6. However, if d is a power of a
prime number, then the answer turns out to be yes, and in fact we
can say a great deal more. This follows from the Sylow theorems,
which we will prove in this section. These theorems give information
about the subgroups of finite groups, and in a sense provide a
partial converse to Lagrange's theorem.
Sylow Theorems
• p-Groups
4.2.1 THEOREM
Let G be a group of order pn and let X be a finite G-set. Then
|X| = |XG| (mod p).
PROOF
In the notation of Eq. (2), we know that |Gxi| divides |G| by
Theorem 3.5.16. Consequently p divides |Gxi| for s+1 <= i <= r.
Equation (2) then shows that |X| - |XG| is divisible by p, so
|X| = |XG| (mod p)
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.
Sylow Theorems
• p-Groups
Cauchy's Theorem.
Let G be a finite abelian group and p a prime that divides |G|. Then
G has an element of order p.
Proof (by induction on order of G).
If |G| = 1, the theorem is obviously true. Let |G| = n, and assume the
theorem holds for all groups with orders less than n. G must have
elements of prime order, since if |x| = qm, where q is prime, |xn| = q.
Let x be an element of G and have order q. Assume q ≠ p, and
constuct <x>. Since every subgroup of an abelian group is
normal, G/<x> is also an abelian group. The order of this group is
n/q, which is divisible by p, and so by the induction hypothesis,
G=<x> contains an element y<x> of order p. Thus (y<x>)p = <x>,
meaning yp is an element of <x>, so either yp = e or |yp| = q, in which
case |yq| = p, since ypq = yqp = e.
Sylow Theorems
• Sylow Theorems
Normalizer
The subgroup GH just discussed is the normalizer of H in G and will
be denoted N[H] from now on
LEMMA
Let H be a p-subgroup of a finite group G. Then
(N[H] : H) = (G : H) (mod p).
PROOF
Consider the set S of all left cosets [G : H] = {gH | g element of G}.
Let H acts on [G : H], by the following operation, h (gH) → (hg)H,
which is well-defined. As before, we have |S| = |S0| (mod p),
where S0 = {gH | hgH = gH for all h element of H}
= {gH | g-1hg element of H for all h element of H}
= {gH | g-1Hg = H} = [NG(H) : H].
Thus, the result follows.
Sylow Theorems
• Sylow Theorems
COROLLARY
Let H be a p-subgroup of a finite group G. If p divides (G : H),
then N[H] ≠ H
PROOF
It follows from LEMMA that p divides (N[H] : H), which must then be
different from 1. Thus H ≠ N[H].
Sylow Theorems
• Sylow Theorems
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.
Sylow Theorems
• Sylow Theorems
Example.
If G=A4, then |G|=12=22.3;
thus a Sylow 2-subgroup has order 4 and a Sylow 3-subgroup has
order 3.
We may take
as examples of each.
Sylow's first theorem tells us that Sylow p-subgroups exist;
his second theorem concerns their conjugacy.
Sylow Theorems
• Heuristic Analysis of Sylow's Theorems