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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.

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2 Answers 2

The historical motivation for defining solvability has to do with Galois theory, namely the theorem which states that the root of a polynomial with rational coefficients are expressible in terms of radicals if and only if the Galois group of the splitting field of the polynomial is solvable.

For a slightly more modern take, at least for finite groups, one might begin with the Jordan–Hölder theorem, which says, among other things, that it is possible to decompose a finite group $G$ into a collection of simple groups in a unique way. (This decomposition is analogous to, but much more complicated than, prime factorization.) The simplest (!) simple groups are the abelian ones, namely the cyclic groups of prime order, and the finite solvable groups are precisely the ones which can be built up from abelian groups in this way. (For infinite groups the Jordan–Hölder theorem fails, but the idea of building up a group from abelian groups is still sound.)

It is a straightforward exercise, and one you should attempt to do on your own, to verify that the only finite solvable simple groups are the cyclic groups of prime order. Since a group with order $p^a q^b$ with $a, b \ge 1$ cannot be cyclic of prime order, Burnside's theorem implies that it cannot be simple.

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Thank you, @Qiaochu, this is very helpful! :) –  Alfonso Catoia Jan 26 '12 at 10:04
    
I tried proving that the only finite solvable simple groups are cyclic and of prime order, but I am not really getting anywhere. I am new to "groups", could you please help me out a bit more? thanks. –  Alfonso Catoia Jan 26 '12 at 11:11

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?

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Thank you! Your definition of a solvable group is more useful that the one on Wikipedia which says that each factor group (which I thought was a $G_i$ in the chain -- like a "factor" in the chain -- instead of the quotient group) is abelian... –  Alfonso Catoia Jan 26 '12 at 12:46
    
@Alfonso, now you have digested all of the above, can you prove that the following two statements are equivalent: (a) every finite simple group must have even order (b) every finite group of odd order is solvable. The (a) is a famous conjecture of Burnside (1911). It has been proved by Feit and Thompson in 1963 (The Odd Order Theorem). The proof is very very long, although it has been substantially simplified over the years. –  Nicky Hekster Jan 27 '12 at 11:56

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