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?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Yes, corrected. $\endgroup$ – Derek Holt Dec 19 '18 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.