Fourth Sylow Theorem/Proof 2

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.

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.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 58$: Third Sylow Theorem