Proof of the second Sylow theorem

Proof

We let a $p$-subgroup act on the cosets of a Sylow subgroup and count fixed points modulo $p$.

1. Write $|G| = p^n m$ with $n \geq 0$ and $p \nmid m$. Each Sylow $p$-subgroup of $G$ has order $p^n$.
2. Let $H$ be a $p$-subgroup of $G$, and let $P$ be a Sylow $p$-subgroup of $G$. Such a $P$ exists by the first Sylow theorem.
3. Let $X = G/P$ be the set of left cosets of $P$. The group $G$ acts on $X$ by $g \cdot aP = gaP$, by the action on left cosets. Restrict this action to $H$.
4. By the index formula, $|X| = |G|/|P| = p^n m / p^n = m$.
5. Let $X^H$ be the set of cosets fixed by every element of $H$. Since $H$ is a finite $p$-group, the fixed point lemma gives $|X^H| \equiv |X| = m \pmod p$.
6. Since $p \nmid m$, we get $|X^H| \not\equiv 0 \pmod p$. So $X^H$ is not empty. Pick a coset $gP \in X^H$.
7. For each $h \in H$ we have $hgP = gP$. So $g^{-1}hg \in P$, that is, $h \in gPg^{-1}$. Hence $H \subseteq gPg^{-1}$.
8. The map $x \mapsto gxg^{-1}$ is a bijection from $P$ onto the conjugate subgroup $gPg^{-1}$. So $gPg^{-1}$ is a subgroup of order $p^n$, a Sylow $p$-subgroup of $G$.
9. By steps 7 and 8, each $p$-subgroup $H$ lies in some Sylow $p$-subgroup. This is the second claim.
10. Now let $P$ and $Q$ be Sylow $p$-subgroups. A Sylow $p$-subgroup has order $p^n$, so the order of each of its elements divides $p^n$ and is a power of $p$. So $Q$ is a $p$-subgroup.
11. Apply steps 3 to 8 with $H = Q$. This gives $g \in G$ with $Q \subseteq gPg^{-1}$.
12. Both $Q$ and $gPg^{-1}$ have $p^n$ elements, and the first is contained in the second. So $Q = gPg^{-1}$. This is the first claim.