Solvability and Simplicity I have read that Burnside's theorem implies that a group with order $p^aq^b$ cannot be simple. So I looked up Burnside's theorem and saw that it doesn't mention "simple" explicitly, rather it says that such a group is solvable. Firstly, I am not sure I fully appreciate/understand the definition of "solvability". What is the motivation behind defining such a property and how does it imply non-simplicity in our $p^aq^b$ group?
Added: Could anyone please explain why "only finite solvable simple groups are the cyclic groups of prime order" (as suggested by Qiaochu) without quoting any particular theorem? This is probably easy to do, but I don't quite see it. Thanks.

I'm not sure you know the definition of solvable. $G$ is solvable means there is a list of groups $G_1,G_2,\dots,G_n$ such that $G_1$ is the one-element group, each $G_i$ is normal in $G_{i+1}$, each quotient group $G_{i+1}/G_i$ is cyclic of prime order, and $G_n$ is $G$. 
Meanwhile, if $G$ is simple, well, that means it has no normal subgroups other than itself and the one-element group, so the only possible list of groups beginning with 1, ending with $G$, and each normal in the next, is the list, $1,G$. So if $G$ is simple and solvable, then the quotient $G/1$ must be cyclic of prime order. OK?
