Let $G=G^{'}Z(G)$ be a finite non-solvable group, $N$ a simple non-abelian subgroup of $G$ such that $Z(G)\leq N$ and $\frac{G}{N}\cong A$ be a non-abelian simple group. Is it true that $G=N\times A$ ?

  • $\begingroup$ Do you mean to assume $N$ to be normal? And do you see that the conditions force $Z(G) = 1$? $\endgroup$ – Tobias Kildetoft Jan 23 '15 at 14:35
  • $\begingroup$ This question, along with your last one, both seem to have a lot of assumptions that are superfluous. Maybe it would help to ask the original question, rather than some small part of it. $\endgroup$ – Tobias Kildetoft Jan 23 '15 at 14:38
  • $\begingroup$ @Tobias Kildetoft: Yes I mean $N$ is a normal subgroup of $G$. I don't know how it implies that $Z(G)=1$ and I didn't get the answer of my question yet. $\endgroup$ – M. R. Jan 23 '15 at 14:44
  • $\begingroup$ Well, clearly anything central in $G$ will centralize $N$, which is assumed non-abelian simple. $\endgroup$ – Tobias Kildetoft Jan 23 '15 at 14:46
  • $\begingroup$ The answer to the question is yes, but the proof uses the Schreier Conjecture, whose proof depends on the classification of finite simple groups. But it should be completely clear that if $Z(G) \le N$ and $N$ is a nonabelian simple group, then $Z(G)=1$. $\endgroup$ – Derek Holt Jan 23 '15 at 15:28

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