What I did: let |G|= p^m q^n and suppose G simple. Let T(g) be the conjugacy class of some element g in G. Of course |Z(G)|= 1. If |T(g)| != 1 then pq | |T(g)|, for T(g) cant be a prime power and T(g) | |G|. Therefor the class equation is |G| = 1 + c, where p|c and hence p | 1, which is absurd. Therefor G has a nontrivial proper normal subgroup H. (1)

If |G|=1 then G is vacuously solvable [?]. Assume the statement in Burnside'a theorem true for every group of order less than |G|. By (1) G has a nontrivial proper normal subgroup H. Now |H| < |G| and |G/H| < |G|. By the induction hypothesis H and G/H are solvable. According to a previous theorem then G is solvable.

The question: is this a valid proof by mathematical induction? For certainly |G| does not take all integer values less than |G|. In other words, shouldn't I do induction on m, or induction on both m and n?