Let $G=G^{'}Z(G)$ be a finite non-solvable group, $N$ an abelian minimal normal subgroup of $G$ ( $|N|=p^d$ for some integer $d$ and prime $p\neq 2,3,5$) such that $N=C_G(N)$, $Z(G)\leq N$ and $\frac{G}{N}\cong A_5$. If $x\in A_5$ is an element of order 3, then how can we show that $C_G(x)=<x>C_N(x)$?

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    $\begingroup$ Technically the question doesn't make sense, because you have defined $x$ as an element of $A_5$ and not of $G$. I guess you mean $x \in G$ and $xN$ has order $3$. The conclusion $C_G(x) = \langle x \rangle C_N(x)$ follows immediately from $C_G(xN) = \langle xN \rangle$, and most of your assumptions are irrelevant. You just need $N \lhd G$ with $G/N \cong A_5$. $\endgroup$ – Derek Holt Jan 23 '15 at 14:24

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