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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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Can you prove/use that the Frattini subgroup of a finite group is nilpotent? – DonAntonio Oct 8 '12 at 10:35
Yes, that has already been proven. – awllower Oct 8 '12 at 10:38
I mean it has been proven by previous exercises... – awllower Oct 8 '12 at 10:45
up vote 3 down vote accepted

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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Thanks very much: I did not come across the part that $K/ \Phi$ is a Sylow subgroup, thus is normal. Thanks for the answer. – awllower Oct 8 '12 at 18:07
By the way, the point you referred to in your comment is not used here. Perhaps you wanted to apply it somehow? – awllower Oct 8 '12 at 18:08
Yes, I know. I wrote below my original comment about this (that this seemed hopeless) but somehow the comment never showed up. – DonAntonio Oct 9 '12 at 3:03

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