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At page (28) of chapter I of the book Finite Group Theory by I.Martin.Issacs, one finds:

Let $G$ be a finite group, with Frattini subgroup $\Phi$. If $\Phi \subseteq N \vartriangleleft G$, and if $N/ \Phi$ is nilpotent, then $N$ is nilpotent.

I know that this is in fact a generalisation of two preceding exercises, but I could not prove it: I have tried to construct a central series of $N$ from that of $N/ \Phi$, but failed to do so. I also tried in the direction of showing that maximal subgroups of $N$ are normal in $N$, but thus far found nothing interesting. As this is a generalisation of one exercise which deploits of the Frattini arguments, I wanted to avail myself of that argument as well, while finding nothing critical either. Therefore I post here for some help.
Sincere thanks.

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    I mean it has been proven by previous exercises...2012-10-08

1 Answers 1

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We know that a finite group $\,G\,$ is nilpotent iff all its Sylow $\,p$-subgroups are normal, so this is what we're going to try to prove for $\,N\,$, so let $\,P\,$ be any Sylow $\,p$-subgroup of $\,N\,$ .

Putting $\,\Phi(G):=\Phi\,$ for simplicity, define $\,K:=P\Phi\leq N\,$ . Then $\,K/\Phi\,$ is a Sylow $\,p$-subgroup of $\,N/\Phi\,$ and since the last group is nilpotent then $\,K/\Phi\,\operatorname{char}\,N/\Phi\triangleleft G/\Phi\Longleftrightarrow K\triangleleft G\,$ (Remember that a finite group is nilpotent iff all its Sylow subgroups are normal, and in fact, characteristic).

Now use Frattini's Argument: $G=N_G(P)K=N_G(P)\Phi\Longrightarrow G=N_G(P)\Longleftrightarrow P\triangleleft G\Longrightarrow P\triangleleft N$

and were done.

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    Yes, I know. I wrote below my original comment about this (that this seemed hopeless) but somehow the comment never showed up.2012-10-09