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In this question, the OP mentions the following result: a finite group $G$ is solvable if and only if

$$\text{for all $n$ dividing $|G|$ such that $\gcd(\frac{|G|}{n},n)=1$, $G$ has a subroup order $n$}. \,\, (*)$$

This was surprising to me: for one thing, this together with the Sylow theorems implies the Burnside Theorem. The OP didn't provide a reference, and a proof wasn't apparent to me. A cursory web search and a perusal of my algebra texts yielded no results.

On the other hand, a counterexample didn't immediately come to mind -- but it would suffice to find a group of order less than $60$ which does not satisfy $(*)$, or to find a finite non-abelian simple group satisfying $(*)$.

My questions is this: is this characterization of solvable groups valid? I'd like either a proof, a reference to a proof, or a counterexample. Many thanks in advance!

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Hall subgroups and Hall' Theorem are described here http://en.wikipedia.org/wiki/Hall_subgroup. The converse of Hall's Theorem uses Burnsides' Theorem.

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    $\begingroup$ This theorem is proven in [Kurzweil, Stellmacher - The Theory of Finite Groups]. See Theorem 6.4.10. $\endgroup$ – Dune Jul 31 '14 at 14:49

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