Burnside's $p^a{q}^b$ theorem

Suppose the statement false, and choose a counterexample $G$ of minimal order. If one of $a$ or $b$ is zero, then $G$ is a finite $r$-group for $r=p$ or $r=q$. If $G$ is trivial or abelian, then its derived subgroup is trivial, so $G$ is solvable by [F7]. Otherwise [F2], with $r$ in place of its generic prime, applied to $G⊴G$ gives a nontrivial central subgroup $N⊊G$. Then both $N$ and $G\mathrm{/}N$ are smaller finite $r$-groups, so minimality makes them solvable, and [F3] makes $G$ solvable, contradiction. So any minimal counterexample has $a,b\ge 1$.

If $N⊴G$ is proper and nontrivial, then both $N$ and $G\mathrm{/}N$ have smaller order of the form $p^r{q}^s$, so minimality makes them solvable; then [F3] makes $G$ solvable, contradiction. Hence a minimal counterexample must be simple.

By [F5], choose a Sylow $p$-subgroup $P\le G$. Since $P$ is a nontrivial finite $p$-group, [F2] applied to $P⊴P$ gives a nonidentity element $z\in Z(P)$. Because $G$ is simple and $a,b\ge 1$, it is not abelian, so $z\mathrm{\notin }Z(G)$.

Every element of $P$ commutes with $z$, so $P\le C_G(z)$. Therefore the full $p$-part of $\mathrm{\mid }G\mathrm{\mid }$ lies in $\mathrm{\mid }{C}_G(z)\mathrm{\mid }$, and [F6] gives $\mathrm{\mid }{\mathrm{Cl}}_G(z)\mathrm{\mid }=[G:C_G(z)]=q^m$ for some $m\ge 1$. Applying [F1] to this prime-power class yields a proper nontrivial normal subgroup of $G$, contradicting simplicity from step 2.1.

The contradiction shows that no counterexample exists. Therefore every finite group of order $p^a{q}^b$ is solvable.