Largest normal soluble subgroup.

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!

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

Take $\;G=S_3\;,\;\;N:=A_3$ . Then $\;S(G)=G\;$ and $\;G/N\cong C_3\;$ has no non-trivial subgroups whatsoever.
• What about $G/N$ itself ? – mesel Mar 12 '17 at 11:19
• @DonAntonio, thank you. But with the condition that $G/N$ is non-abelian, my conclusion is correct? – Sibilla Mar 12 '17 at 11:21
• Trivial subgroup of $G/N$ is only identity subgroup. Am I missing something ? – mesel Mar 12 '17 at 11:21