Fourth Sylow Theorem/Proof 2
Let $r$ be the number of Sylow $p$-subgroups of $G$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
- $\order H = p^n$
- $\index G H = m$
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.
From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases:
So counting the elements of $G / H$, we see that:
- $m = k + u p$
- $m \equiv k \pmod p$
From the Fifth Sylow Theorem, we have:
- $m \equiv k r \pmod p$
- $k r \equiv k \pmod p$
from which it follows:
- $r \equiv 1 \pmod p$
because $k \not \equiv 0 \pmod p$.
Hence the result.