Third Sylow Theorem/Proof 2

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

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 left cosets of $G \pmod H$.

We have that $G$ acts on $G / H$ by the rule:
 * $g * S_i = g S_i$.

Let $H_i$ denote the stabilizer of $S_i$.

By the Orbit-Stabilizer Theorem:
 * $\order {H_i} = p^n$

while:
 * $S_i = g H \implies g H g^{-1} \subseteq H_i$

Because $\order {g H g^{-1} } = \order H = \order {H_i}$, we have:
 * $g H g^{-1} \subseteq H_i$

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

Then $H'$ acts on $G / H$ by the same rule as $G$.

Since $p \nmid m$, there exists at least one orbit under $H'$ whose cardinality is not divisible by $p$.

Suppose that $S_1, S_2, \ldots, S_r$ are the elements of an orbit where $p \nmid r$.

Let $K = H' \cap H_1$.

Then $K$ is the stabilizer of $S_1$ under the action of $H'$.

Therefore:
 * $\index {H'} K = r$

However:
 * $\order {H'} = p^n$

and:
 * $p \nmid r$

from which it follows that:
 * $r = 1$

and:
 * $K = H'$

Therefore:
 * $\order K = \order {H'} = \order {H_1} = p^n$

and:
 * $H' = K = H_1$

Thus $H'$ and $H$ are conjugates.