Explaining Finite Solvable Groups: Understanding Burnside's Theorem

[MarkovMarakov]

HI, I was reading an article and it says that a finite group of order $p^aq^b$, where p, q are primes, is solvable and therefore not simple. But I can't quite understand why this is so. I do recall a theorem called Burnside's theorem which says that a group of such order is solvable. But then I don't see how it follows that the group is simple. Could someone please explain? Thanks.

 

[micromass]

A simple solvable group has to be of prime order. Indeed, a simple group does not have any normal subgroups, so the only subnormal series has to be $\{1\}\leq G$. But solvability says that the quotient needs to be abelian. This means that G is abelian and simple and means that it's isomorphic to $\mathbb{Z}_p$.

So a group of order $p^aq^b$ with a,b>0 has to be solvable. If it were simple then it had to be of prime order. But this cannot be since both a,b>0.

 

[MarkovMarakov]

@micromass: Thanks!