# 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.

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?

## 1 Answer

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.

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