Proof of the third Sylow theorem

Proof

We let a Sylow subgroup act by conjugation on the set of all Sylow subgroups and count fixed points modulo $p$.

1. Write $|G| = p^n m$ with $n \geq 0$ and $p \nmid m$. Let $\Omega = \operatorname{Syl}_p(G)$ be the set of Sylow $p$-subgroups, so $|\Omega| = n_p$. By the first Sylow theorem, $\Omega$ contains some $P$.
2. For $g \in G$ and $Q \in \Omega$, the conjugate subgroup $gQg^{-1}$ is a subgroup of order $|Q| = p^n$, since $x \mapsto gxg^{-1}$ is a bijection. So $gQg^{-1} \in \Omega$.
3. So $G$ acts on $\Omega$ by $g \cdot Q = gQg^{-1}$, as in the action by conjugation. Restrict this action to $P$.
4. The order of each element of $P$ divides $p^n$, so it is a power of $p$. So $P$ is a finite $p$-group. The fixed point lemma gives $n_p = |\Omega| \equiv |\Omega^P| \pmod p$, where $\Omega^P$ is the set of fixed points.
5. The subgroup $P$ is a fixed point, because $xPx^{-1} = P$ for each $x \in P$.
6. Let $Q \in \Omega^P$. Then $xQx^{-1} = Q$ for each $x \in P$. So $P \subseteq N$, where $N = N_G(Q)$ is the normaliser of $Q$ in $G$.
7. The set $N$ is a subgroup of $G$. Indeed, if $gQg^{-1} = Q$ and $hQh^{-1} = Q$, then $(gh)Q(gh)^{-1} = Q$, and $g^{-1}Qg = Q$ follows from $Q = gQg^{-1}$.
8. Also $Q \subseteq N$, and $Q$ is normal in $N$ by the definition of $N$.
9. By Lagrange's theorem, $|N|$ divides $p^n m$, and $p^n = |Q|$ divides $|N|$. So $|N| = p^n m'$ with $m' \mid m$, hence $p \nmid m'$.
10. So $P$ and $Q$ are both subgroups of $N$ of order $p^n$, that is, Sylow $p$-subgroups of $N$.
11. By the second Sylow theorem in the group $N$, there is $g \in N$ with $P = gQg^{-1}$.
12. Since $g \in N = N_G(Q)$, we have $gQg^{-1} = Q$. So $P = Q$.
13. By steps 5 and 12, $\Omega^P = \{P\}$, so $|\Omega^P| = 1$.
14. By step 4, $n_p \equiv 1 \pmod p$.