Proof of the first Sylow theorem by induction with the class equation

Proof

We argue by strong induction on $\abs{G}$. Assume that every group of order less than $\abs{G}$ has a subgroup of order $p^j$ whenever $p^j$ divides its order. Let $e$ be the identity of $G$.

1. If $k = 0$, the subgroup $\set{e}$ has order $p^0 = 1$. So assume $k \geq 1$; then $p$ divides $\abs{G}$. We split into two cases.
2. Case 1: $p$ divides $\abs{Z(G)}$. The centre $Z(G)$ is a subgroup of $G$, by the subgroup criterion. Indeed $e \in Z(G)$. If $u, w \in Z(G)$ and $g \in G$, then $w g = g w$ gives $w^{-1} g = g w^{-1}$, so $u w^{-1} g = u g w^{-1} = g u w^{-1}$. By Cauchy's theorem, it has an element $z$ of order $p$.
3. Let $N = \langle z \rangle$. By the order of a cyclic subgroup, $N = \set{e, z, \dots, z^{p-1}}$ and $\abs{N} = p$.
4. For $g \in G$, each $z^i$ commutes with $g$, so $g z^i g^{-1} = z^i$. Hence $g N g^{-1} = N$, and $N$ is a normal subgroup of $G$.
5. Let $\bar G = G/N$ be the quotient group. By the index formula, $\abs{\bar G} = [G : N] = \abs{G}/p$. This is less than $\abs{G}$ and divisible by $p^{k-1}$.
6. By the induction hypothesis, $\bar G$ has a subgroup $\bar H$ of order $p^{k-1}$.
7. Let $\pi \colon G \to \bar G$, $\pi(g) = gN$. By the definition of the quotient operation, $\pi(gh) = \pi(g)\pi(h)$, and $\pi(g^{-1}) = \pi(g)^{-1}$.
8. Let $H = \set{g \in G : \pi(g) \in \bar H}$. It contains $e$. If $g, h \in H$, then $\pi(g h^{-1}) = \pi(g) \pi(h)^{-1} \in \bar H$. So $H$ is a subgroup of $G$.
9. Each element of $\bar H$ is a coset $aN$ with $a \in G$. For $x \in aN$, write $x = an$ with $n \in N$; then $xN = a(nN) = aN$, because $m \mapsto nm$ is a bijection of $N$.
10. By step 9, $x \in H$ if and only if $x$ lies in some coset belonging to $\bar H$. So $H$ is the union of the $p^{k-1}$ cosets in $\bar H$.
11. Distinct cosets are disjoint: if $x \in aN \cap bN$, step 9 gives $aN = xN = bN$. Each coset has $\abs{N} = p$ elements.
12. By steps 10 and 11, $\abs{H} = p^{k-1} \cdot p = p^k$. This settles Case 1.
13. Case 2: $p$ does not divide $\abs{Z(G)}$. By the class equation, $\abs{G} = \abs{Z(G)} + \sum_{i=1}^{r} [G : C_G(a_i)]$, over the conjugacy classes with more than one element.
14. If $p$ divided every $[G : C_G(a_i)]$, it would divide $\abs{G} - \sum_i [G : C_G(a_i)] = \abs{Z(G)}$. So some $i$ has $p \nmid [G : C]$, where $C = C_G(a_i)$.
15. The centraliser $C$ is a subgroup of $G$. Indeed $e \in C$. If $g, h \in C$, then $h^{-1} a_i = a_i h^{-1}$, so $g h^{-1} a_i = g a_i h^{-1} = a_i g h^{-1}$. By the class size formula, $[G : C] = \abs{\operatorname{Cl}(a_i)} > 1$.
16. By the index formula, $\abs{G} = [G : C] \cdot \abs{C}$. With step 15, $\abs{C} < \abs{G}$.
17. A common prime factor of $p^k$ and $[G : C]$ would divide $p$ by Euclid's lemma, so it would be $p$. This is excluded, so $p^k$ and $[G : C]$ are coprime.
18. Since $p^k$ divides $[G : C] \cdot \abs{C}$, the generalised Euclid's lemma gives $p^k \mid \abs{C}$.
19. By steps 16 and 18 and the induction hypothesis, $C$ has a subgroup $K$ of order $p^k$. Then $K$ is also a subgroup of $G$. This settles Case 2.
20. By induction, the statement holds for every finite group $G$ and every $k \geq 0$ with $p^k \mid \abs{G}$.
21. Write $\abs{G} = p^n m$ with $p \nmid m$. Taking $k = n$ gives a subgroup of order $p^n$, which is a Sylow $p$-subgroup.