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How does one prove that a group of order $p^xq^y$ where $p,q$ are distinct primes and $x,y>1$ are integers, is NOT simple, without using Burnside's theorem or solvability?

I think one way to proceed is to observe that by Sylow's third theorem, if $n_p$ is the number of Sylow $p-$subgroups then $n_p|p^xq^y$ and $n_p\equiv 1\pmod p$ so that $n_p\in\{1,q,q^2,...,q^y\}$. Suppose $n_p=q^m$ for some $1\leq m\leq y$ but then I cannot conclude that there are $n_p-1$ many elements of order $p^x$ since there can be non-trivial intersections...

Landon Carter
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You can't, because the fact that such groups are not simple is effectively equivalent to the fact that they are solvable.

If they are solvable then they have a normal series with abelian factors and so are not simple.

Conversely, it is easy to see by induction that they are solvable. If such a group is not simple, then it has a proper nontrivial normal subgroup $N$. Then each of $N$ and $G/N$ is either a $p$-group, in which case it is solvable, or it is solvable by induction. So $N$ and $G/N$ are both solvable, and hence so is $G$.

Derek Holt
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