Fourth Sylow Theorem

Theorem
The number of Sylow $p$-subgroups of a finite group is congruent to $1 \left({\bmod p}\right)$.

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.

Proof
Let $G$ be a finite group such that $\left|{G}\right| = k p^n$ where $p \nmid k$ and $n > 0$.

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

We want to show that $r \equiv 1 \left({\bmod p}\right)$.


 * Let $\mathbb S = \left\{ {S \subseteq G: \left| {S} \right| = p^n} \right\}$, 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:
 * $\displaystyle \left|{\mathbb S}\right| = \binom {p^n k} {p^n}$

Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements: $\forall S \in \mathbb S: g \wedge S = g S = \left\{{x \in G: x = g s: s \in S}\right\}$.

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 $k$. Some of them may also be divisible by $p$.

We can write the Partition Equation as:


 * $\left|{\mathbb S}\right| = \left|{\operatorname{Orb} \left({S_1}\right)}\right| + \left|{\operatorname{Orb} \left({S_2}\right)}\right| + \cdots + \left|{\operatorname{Orb} \left({S_r}\right)}\right| + \left|{\operatorname{Orb} \left({S_{r+1}}\right)}\right| + \cdots + \left|{\operatorname{Orb} \left({S_s}\right)}\right|$

where the first $r$ terms are the orbits containing the Sylow $p$-subgroups: $\operatorname{Stab} \left({S_i}\right)$.

For each of these:
 * $\left|{G}\right| = \left|{\operatorname{Orb} \left({S_i}\right)}\right| \times \left|{\operatorname{Stab} \left({S_i}\right)}\right| = p^n \left|{\operatorname{Orb} \left({S_i}\right)}\right|$

Thus $\left|{\operatorname{Orb} \left({S_i}\right)}\right| = k$ 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 $\left|{\mathbb S}\right| = k r + m p k$, where the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups, and the second term to all the rest of the orbits, and $m$ is some unspecified integer.

That is, there exists some integer $m$ such that $\displaystyle \left|{\mathbb S}\right| = \binom {p^n k} {p^n} = k r + m p k$.

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 $\displaystyle \binom {p^n k} {p^n} = k + m' p k$.

We can now equate these expressions:

and the proof is complete.