Proof of a Burnside theorem without character theory?

Burnside proved, by use of character theory, that if a finite group $G$ has a conjugacy class $C$ such that $\vert C \vert$ is a prime power $> 1$, then $G$ is not simple. Let us call this statement "Burnside's non-simplicity theorem".

From this theorem, Burnside deduced easily his "$p^{a}q^{b}$ theorem" : every finite group whose order has at most two distinct prime factors is solvable.

It is well known that Burnside's $p^{a}q^{b}$ theorem can be proved indepedently from character theory. Can "Burnside's non-simplicity theorem" also be proved indepedently from character theory ?

• If would help if you'd show a sketch proof using character theory – reuns Dec 1 '17 at 19:07
• Sorry, for a moment I thought your question was "Can Burnside's non-simplicity Theorem also be proved independently from Group Theory"... – Dietrich Burde Dec 1 '17 at 19:20
• @reuns : I think that any person who knows the first elements of character theory knows the proof of "Burnside's non-simplicity theorem" that uses character theory. I never saw an introduction to character theory that doesn't give this proof. – Panurge Dec 2 '17 at 7:29
• As you pointed out, Burnside's $p^aq^b$ can easily be deduced from what you call "Burnside's non-simplicity theorem", and this without using any character theory. So you essentially ask for a purely group-theoretic proof of Burnside's $p^aq^b$-theorem. If the known such proofs by Bender rsp. Goldschmidt and Matsuyama cannot be used for proving "Burnside's non-simplicity theorem", then it is unlikely that someone will be able to answer your question. – j.p. Dec 2 '17 at 17:59
• @reuns : I only wish to know if the literature contains a proof of "Burnside's non-simplicity theorem" without character theory. I don't expect that a member of MSE will invent such a proof. – Panurge Dec 3 '17 at 7:21