# Fourth Sylow Theorem

## Contents

## Theorem

The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$.

## Proof 1

Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$ and $n > 0$.

Let $r$ be the number of Sylow $p$-subgroups of $G$.

We want to show that $r \equiv 1 \pmod p$.

Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.

From the reasoning in the First Sylow Theorem, we have:

- $\size {\mathbb S} = \dbinom {p^n k} {p^n}$

Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements::

- $\forall S \in \mathbb S: g \wedge S = g S = \set {x \in G: x = g s: s \in S}$

From Orbits of Group Action on Sets with Power of Prime Size:

- there are exactly as many Sylow $p$-subgroups as there are orbits whose length is not divisible by $p$.

Also by Orbits of Group Action on Sets with Power of Prime Size:

- all the terms in the Partition Equation are divisible by $k$, perhaps also divisible by $p$.

We can write the Partition Equation as:

- $\size {\mathbb S} = \size {\Orb {S_1} } + \size {\Orb {S_2} } + \cdots + \size {\Orb {S_r} } + \size {\Orb {S_{r + 1} } } + \cdots + \size {\Orb {S_s} }$

where the first $r$ terms are the orbits containing the Sylow $p$-subgroups:

- $\Stab {S_i}$

For each of these:

- $\order G = \size {\Orb {S_i} } \times \size {\Stab {S_i} } = p^n \size {\Orb {S_i} }$

Thus:

- $\size {\Orb {S_i} } = k$

for $1 \le i \le r$.

Each of the rest of the orbits are divisible by both $p$ and $k$, as we have seen.

So:

- $\size {\mathbb S} = k r + m p k$

where:

- the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups
- the second term corresponds to all the rest of the orbits
- $m$ is some unspecified integer.

That is, there exists some integer $m$ such that:

- $\size {\mathbb S} = \dbinom {p^n k} {p^n} = k r + m p k$

Now this of course applies to the special case of the cyclic group $C_{p^n k}$.

In this case, there is exactly one subgroup for each divisor of $p^n k$.

In particular, there is exactly one subgroup of order $p^n$.

Hence, in this case:

- $r = 1$

So we have an integer $m'$ such that $\dbinom {p^n k} {p^n} = k + m' p k$.

We can now equate these expressions:

\(\displaystyle k r + m p k\) | \(=\) | \(\displaystyle k + m' p k\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r + m p\) | \(=\) | \(\displaystyle 1 + m' p\) | dividing by $k$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r - 1\) | \(=\) | \(\displaystyle p \paren {m' - m}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r - 1\) | \(\equiv\) | \(\displaystyle 0 \pmod p\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r\) | \(\equiv\) | \(\displaystyle 1 \pmod p\) |

and the proof is complete.

$\blacksquare$

## Proof 2

Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.

Let $r$ be the number of Sylow $p$-subgroups of $G$.

Let $H$ be a Sylow $p$-subgroup of $G$.

We have that:

- $\order H = p^n$

- $\index G H = m$

Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.

We have that $H$ acts on $G / H$ by the rule:

- $g * S_i = g S_i$

for $S_i \in G / H$.

Unless $H = G$ and $r = 1$, there is more than $1$ orbit.

We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$.

Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$.

From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases:

$(2)$ occurs if and only if $S_i$ is one of the cosets $S_1, S_2, \ldots, S_k$ whose stabilizer is $H$.

So counting the elements of $G / H$, we see that:

- $m = k + u p$

or:

- $m \equiv k \pmod p$

From the Fifth Sylow Theorem, we have:

- $m \equiv k r \pmod p$

and so:

- $k r \equiv k \pmod p$

from which it follows:

- $r \equiv 1 \pmod p$

because $k \not \equiv 0 \pmod p$.

Hence the result.

$\blacksquare$

## Also known as

Some sources call this the **second Sylow theorem**.

Others merge this result with what we call the Fifth Sylow Theorem and call it the **third Sylow theorem**.

## Also see

## Source of Name

This entry was named for Peter Ludwig Mejdell Sylow.

## Historical Note

When cracking open the structure of a group, it is a useful plan to start with investigating the prime subgroups.

The **Sylow Theorems** are a set of results which provide us with just the sort of information we need.

Ludwig Sylow was a Norwegian mathematician who established some important facts on this subject.

He published what are now referred to as the **Sylow Theorems** in $1872$.

The name is pronounced something like **Soolof**.

There is no standard numbering for the **Sylow Theorems**.

Different authors use different labellings.

Therefore, the nomenclature as defined on $\mathsf{Pr} \infty \mathsf{fWiki}$ is to a greater or lesser extent arbitrary.