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Let G be a non-solvable group and $(\frac{G}{Z(G)})^{'}=\frac{G}{Z(G)}$. Can we say that there is a normal subgroup $N$ of $G$ with property $Z(G)\leq N$ such that $\frac{G}{N}$ is a non-abelian simple group ?

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Choose $N \supseteq Z(G)$ maximal normal. Then obviously $G/N$ is simple. Assume $G/N$ is abelian, then $G' \subseteq N$. Since $(G/Z(G))'=G/Z(G)$, it follows that $G=G'Z(G)$. But then $G=N$, since $N$ contains $Z(G)$ and $G'$. $N$ was maximal and hence proper in $G$, a contradiction, so $G/N$ is non-abelian.

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  • $\begingroup$ You are welcome! Please, asd a common habit on StackExchange, if you deem it as the one your were looking for, tick it as such. Thanks. $\endgroup$ – Nicky Hekster Sep 9 '14 at 13:45

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