Fifth Sylow Theorem/Proof 2

Proof
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 $G$.

It is to be shown that $r \divides m$.

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 $G$ acts on $G / H$ by the rule:
 * $g * S = g S$

for $S \in G / H$.

There is only one orbit under this action, namely the whole of $G / H$.

Therefore the stabilizer of each $S_i$ is a subgroup of $G$ of index $m$ and order $p^n$.

In other words, each $S_i$ has a Sylow $p$-subgroup as a stabilizer.

Now it is shown that each Sylow $p$-subgroup is the stabilizer of one or more of the cosets $S_1, S_2, \ldots, S_m$.

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 $H$.

By the Third Sylow Theorem, any other Sylow $p$-subgroup of $H$ is a conjugates $g H g^{-1}$ of $H$.

Thus it is seen that $g H g^{-1}$ is a stabilizer of the cosets $g S_1, g S_2, \ldots, g S_k$.

So each of the $r$ distinct Sylow $p$-subgroups of $G$ is the stabilizer of exactly $k$ elements of $G / H$.

Thus:
 * $m = k r$

and so:
 * $r \divides m$

as required.