a group $G$ whose order has exactly two prime divisors is not simple, Burnside theorem?

Question

I have to show that a group $G$ whose order has exactly two prime divisors is not simple. I was thinking to use the Burnside theorem which sais that if $\vert G \vert = p^aq^b$ where $p,q$ are primes and $a,b \in \mathbb{Z}_{\ge 0}$ then $G$ is solvable. But I don't know how to continue. Couldn't it be that $\{1\}$ is the only normal subgroup of $G$?

Answer 1

By Burnside, you find $H\lhd G$ with $G/H$ cyclic. If $H=1$, this means $G$ itself is cyclic. The only abelian simple groups are cyclic of prime order.