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I want to know a proof of this fact: "every simple group has a minimal simple group as a subquotient." (If $H$ and $K$ are two subgroups of $G$ s.t. $H\lhd K$, then $\frac{K}{H}$ is called a subquotient of $G$) many thanks.

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If every proper subgroup of G is soluble, then it is done. So there is a non-soluble subgroup H with every subgroup soluble. Let K be maximal normal subgroup of H, then H/K is a minimal simple group.

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So do not need $G$ itself to be simple. It is enough that $G$ is not solvable (allowing the possibility $H = G$). –  Geoff Robinson Jan 2 '13 at 8:01
    
Thanks a lot for the answer and correction. –  liobei Jan 2 '13 at 9:06

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