Why is a group of order p^aq^b solvable by Burnside's theorem?

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MarkovMarakov
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HI, I was reading an article and it says that a finite group of order [itex]p^aq^b[/itex], 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.
 
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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 [itex]\{1\}\leq G[/itex]. 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 [itex]\mathbb{Z}_p[/itex].

So a group of order [itex]p^aq^b[/itex] 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.