Fourth Sylow Theorem/Proof 2

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Theorem

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


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 $G$.


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.



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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:

$(1): \quad$ The orbit of $S_i$ contains $p^t$ elements where $0 < t < n$
$(2): \quad$ The orbit of $S_i$ contains only the element $S_i$.


$(2)$ 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.

$\blacksquare$



Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Much of the first part of Fifth Sylow Theorem/Proof 2 can be extracted as a lemma, as it is used as the initial basis of this proof.
Examine this proof carefully and see whether there are points of commonality with the other Sylow theorems so as to genuinely extract some building blocks.
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Sources

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