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Let $G$ a finite group and suppose that $S(G)$$\ne$$1$ where $S(G)$ denotes the largest normal soluble subgroup of $G$. Suppose that $N$ is non-trivial normal soluble subgroup $N$$\ne$$1$, $N$$\vartriangleleft$ $G$ such that the quotient group $G$/$N$ has no nontrivial abelian normal subgroups. Can I conclude that $S(G)$$=$$N$ ? Is it correct? Thank you very much to everyone for the help!

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    $\begingroup$ Yes, you can. Where to Stack ? it is quite elementary. Just Check the image of $S(G)$. $\endgroup$ – mesel Mar 12 '17 at 11:15
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    $\begingroup$ Also there is no need to assume that $G$ is finite. $\endgroup$ – Derek Holt Mar 12 '17 at 11:32
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Counterexample:

Take $\;G=S_3\;,\;\;N:=A_3$ . Then $\;S(G)=G\;$ and $\;G/N\cong C_3\;$ has no non-trivial subgroups whatsoever.

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  • $\begingroup$ What about $G/N$ itself ? $\endgroup$ – mesel Mar 12 '17 at 11:19
  • $\begingroup$ @mesel It is written there "no nontrivial normal..." I tried to go exactly by what the OP wrote. $\endgroup$ – DonAntonio Mar 12 '17 at 11:20
  • $\begingroup$ @DonAntonio, thank you. But with the condition that $G/N$ is non-abelian, my conclusion is correct? $\endgroup$ – Sibilla Mar 12 '17 at 11:21
  • $\begingroup$ Trivial subgroup of $G/N$ is only identity subgroup. Am I missing something ? $\endgroup$ – mesel Mar 12 '17 at 11:21
  • $\begingroup$ @Sibilla Not sure. I'll have to check, though mesel probably was thinking something of the like as he commented below your question that "Yes, you can." Maybe he can post an answer with this assumption $\endgroup$ – DonAntonio Mar 12 '17 at 11:22

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