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Let $G$ be a finite group and $M,N \lhd G$ such that $M \leq N\cap \Phi(G)$. Then $\frac{N}{M}$ is nilpotent iff $N$ is nilpotent, where $\Phi(G)$ is the Frattini subgroup of $G$.

The converse side is obvious. For first side I want to use the below fact.

"Let $G$ be a finite group and $N \lhd G$ such that $N\leq \Phi(G)$. Then $\frac{G}{N}$ is nilpotent iff $G$ is nilpotent."

Let $\frac{N}{M}$ be nilpotent. We have $M \lhd N$. I want to show $M\leq \Phi(N)$. If I show this we have the result from the fact above.

But I don't know how to show it or is it true at all?

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Here is a quick sketch proof. Assume that $N/M$ is nilpotent with $M \le \Phi(G)$.

Let $P \in {\rm Syl}_p(N)$. By the Frattini Argument, $G = N_G(P)N$. We claim that $N_G(P) = G$, which will prove that $N$ is nilpotent.

If not, then let $H$ be a maximal subgroup of $G$ containing $N_G(P)$. Then $M \le \Phi(G) \Rightarrow M \le H$. Now $H \cap N$ contains $N_N(P)$, and so by a well know result $N_N(H \cap N) = H \cap N$, and so $N_{\frac{N}{M}}((H \cap N)/M) = (H \cap N)/M$.

But, $N/M$ is nilptotent, so all of its proper subgroups are strictly contained in their normalizers, and hence $(H \cap N)/M = N/M$ and $N \le H$, so $G=H$, contradiction.

So $N_G(P)=G$ as claimed.

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  • $\begingroup$ thank you very much. $\endgroup$ – Yasmin Dec 19 '18 at 9:19
  • $\begingroup$ In last line you mean $N_{G}(P)= G$ $\endgroup$ – Yasmin Dec 19 '18 at 9:25
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    $\begingroup$ Yes, corrected. $\endgroup$ – Derek Holt Dec 19 '18 at 9:37

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