Sylow’s theorem is a major tool in the project of determining which subgroups of a given group are normal. Relatedly, it represents a significant tool for classifying groups.

Sylow’s theorem regards subgroups that are of order a maximal power of a prime. The details will be explained below. Because I dislike using people’s names to name theorems, I will call it the “p-subgroup” theorem.

Before that we will prove the following theorem, which is interesting in its own right, and is a lemma to Sylow’s theorem.

Theorem (Prime order divisor, the commutative case):

If G is a finite commutative group and p a prime number which divides the order $|G|=n$, then there is a subgroup $H\le G$ with order $p=|H|$.

Proof:

The proof is by strong induction on the order of G.

Exercise:

What is the base case?

After you identify it, prove the base case.

Now suppose that the claim holds up to any n, and consider $|G|=n+1$. That is to say, the inductive hypothesis assumes that for every group of order i, with $1\le i\le n$, if we have $p|i$ then the group has a subgroup of order p.

Note that because the base case was $|G|=p$, then in the inductive case, we have $|G|>p$. This will be useful later.

Because $|G|>1$ there must be a non-identity element $1\ne x\in G$.

Exercise:

Show that if p divides $|x|$, then there is an element of order p.

Hint: This does not require any information from the induction. It has a short proof using only a very important theorem that we’ve used several times.

We therefore only have to consider the case that p does not divide $|x|$. Note that because G is commutative, every one of its subgroups is normal.

Therefore $G/\langle x\rangle$ is a group with order $|G|/|x| < |G|$. Hence the inductive hypothesis applies, and $G/\langle x\rangle$ has an element of order p, call it $y\langle x\rangle$.

Now $y\not\in \langle x\rangle$ but $y^p\in \langle x\rangle$.

Exercise:

Show that $\langle y^p\rangle < \langle y\rangle$.

Since

$$ |y^p| = \frac{|y|}{(|y|,p)} < |y| $$

we must have that $(|y|,p)$ is a multiple of p greater than 1. Therefore p divides $|y|$.

Exercise:

Infer from the inductive hypothesis that y has an element of order p and conclude the theorem.

$\Box$

Definition:

Let G be a group and p a prime number.

If $|G|=p^\alpha$ for some $\alpha\in\Bbb Z^+$, we call G a p-group.

If $H\le G$ is any subgroup, and $|H|=p^\alpha$ for some $\alpha\in\Bbb Z^+$, then we call H a p-subgroup of G.

Definition:

Let $n\in\Bbb Z$ be any integer and p a prime number such that $p|n$.

We call $\alpha\in\Bbb Z^+$ the maximal power for p dividing n if

Let G be a finite group, p a prime number dividing the order of G, and $\alpha$ the maximal power for p dividing $n=|G|$.

If $H\le G$ is any subgroup of order $|H|=p^\alpha$ then H is called a maximal p-subgroup of G.

For each prime p dividing $|G|$, the set of all maximal p-subgroups is denoted $\text{MPS}_p(G)$.

Exercise:

Let p be a prime number dividing $n\in\Bbb Z$. Let $\alpha$ be the maximal power for p dividing n.

Define $m=\frac{n}{p^\alpha}$.

Prove that $p \not | \ \ m$.

Theorem:

Let G be a finite group of order $n=|G|$. Let p be a prime dividing n, $\alpha$ the maximal power of p dividing n, and $n=p^\alpha m$.

  1. There exists a maximal p-subgroup.

  2. Any two maximal p-subgroups are conjugate.

  3. Let $n_p$ denote the number of maximal p-subgroups. Then

    $$ n_p \equiv 1 \mod p $$

    Moreover, for each $P\in \text{MPS}_p(G)$, the number $n_p$ is the index of the normalizer. That is to say,

    $$ n_p = [G:N_G(P)] $$

    Note that this implies $n_p|G$, and so $n_p|m$.

Proof of (1), existence:

The proof is by strong induction on n. If $n=1$ there is nothing to prove.

Suppose that the claim holds up to n, and let $|G|=n+1$. That is to say, we assume that there is a Sylow p-subgroup for each group of order i, with $1\le i\le n$.

By the class equation,

$$ |G| = |Z(G)|+\sum_{j=1}^r [G:C_G(g_i)] $$

where $g_1,\dots,g_r$ are any distinct representatives of the non-central conjugacy classes.

Case 1: p divides $[G:C_G(g_j)]$ for every $1\le j\le r$.

In this case, since we also have that p divides $|G|$, then we have that p divides $|Z(G)|$. By the “prime order divisor, commutative case”, we obtain an element $x\in Z(G)$ of order p.

Case 2: p does not divide $[G:C_G(g_j)]$, for $1\le j\le r$.

Since subgroup order divides group order,

$$ |G| = p^\alpha m = |C_G(g_j)|[G:C_G(g_j)] $$

Since $p{\not|} \ [G:C_G(g_j)]$ therefore $p^\alpha$ must divide $|C_G(g_i)|$.

Since $g_j\notin Z(G)$ we must have that $C_G(g_j) < G$, and therefore the inductive hypothesis applies. Hence there is a subgroup $P\le C_G(g_j)$ with order $p^\alpha$.

This concludes the inductive case, and therefore the proof.

$\Box$

Proof of (2), conjugacy:

Let P be a maximal p-subgroup of G, which we know exists from part (1). Let the set of all conjugates of P be

$$ \mathcal S = \{P_1, P_2, \dots, P_r\} $$

where these are distinct.

Our goal is to prove that the set of maximal p-subgroups is exactly $\mathcal S$. We have seen that multiplication (and therefore also conjugation) does not change the size of a subset. Hence every $P_i$ is a maximal p-subgroup. We merely need to prove the converse.

Let $Q\le G$ be any maximal p-subgroup of G.

Exercise:

Show that G acts by conjugation on $\mathcal S$.

Infer that Q also acts by conjugation on $\mathcal S$.

Write the distinct orbits of $\mathcal S$ as $\mathcal O_1, \mathcal O_2, \dots, \mathcal O_s$. Note that $r=|\mathcal O_1|+\cdots +|\mathcal O_s|$.

Also note that r is independent of Q but s is not.

Since orbits partition a space, each orbit $\mathcal O_i$ must have some representative in $\mathcal S$, for $1\le i\le s$. Let us renumber the elements of $\mathcal S$ so that $P_i\in \mathcal O_i$, for $1\le i\le s$.

We will use the following fact.

$$ |\mathcal O_i| = [Q:P_i\cap Q] $$

The proof of this equation will be given after the end of our current proof.