Third Sylow Theorem/Proof 2

Theorem

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 $S_i$.

$\order {H_i} = p^n$

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.

1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 57$: Second Sylow Theorem