# Burnside’s Theorem

Proofs of this theorem are quite difficult and rely on representation theory.

From this we immediately get

###### Corollary 1.

A group $G$ of order $p^{a}q^{b}$, where $p,q$ are prime, cannot be a nonabelian  simple group.

###### Proof.

Suppose it is. Then the center of $G$ is trivial, $\{e\}$, since the center is a normal subgroup  and $G$ is simple nonabelian. So if $C_{i}$ are the nontrivial conjugacy classes, we have from the class equation   that

 $\lvert G\rvert=1+\sum\lvert C_{i}\rvert$

Now, each $\lvert C_{i}\rvert$ divides $\lvert G\rvert$, but cannot be $1$ since the center is trivial. It cannot be a power of either $p$ or $q$ by Burnside’s theorem. Thus $pq\mid\lvert C_{i}\rvert$ for each $i$ and thus $\lvert G\rvert\equiv 1\pmod{pq}$, which is a contradiction   . ∎

Finally, a corollary of the above is known as the Burnside $p$-$q$ Theorem (http://planetmath.org/BurnsidePQTheorem).

###### Corollary 2.

A group of order $p^{a}q^{b}$ is solvable.

Title Burnside’s Theorem BurnsidesTheorem 2013-03-22 16:38:14 2013-03-22 16:38:14 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 20D05