Order has only two prime factors implies solvable

Statement

The Burnside's p^aq^b theorem, states that if the order of a group has only two prime factors, (viz it is of the form $p^a{q}^b$), then the group must be solvable.

Proof

There exists a conjugacy class of prime power order

Let the order of the group $G$ be $p^a{q}^b$ where $p,q$ are the prime divisors of the order. Let $P$ be a $p$-Sylow subgroup, and let $Z(P)$ be the center of $P$. $Z(P)$ is a nontrivial group. Now for any nonidentity element $g\in Z(P)$, $C_G(g)$ contains $P$, so its index in $G$ is a power of $q$. Hence, the size of its conjugacy class is a power of $q$.

Existence of such a conjugacy class implies solvability

Further information: (Conjugacy class of prime power order) implies not simple

In the page linked to above, it is shown that if such a conjugacy class exists inside the group, then the group is not simple.

=Once we have shown that it is not simple

$G$ now has a proper nontrivial normal subgroup $N$. Both $N$ and $G/N$ have strictly smaller order than $G$ and both of them have one or two prime divisors. Thus, by induction, both $N$ and $G/N$ are solvable groups, and hence, so is $G$.