# Frattini subgroup is normal-monotone

In the exercise 6.1.22 of Dummit and Foote's Abstract Algebra (Here $$\Phi(G)$$ is the Frattini subgroup of $$G$$):

If $$N\unlhd G$$, then $$\Phi(N)\subseteq\Phi(G)$$.

When every proper subgroup of $$N$$ is contained in a maximal subgroup of $$N$$, the statement can be proved. (By taking $$M$$ as a maximal subgroup of $$G$$ that fails to contain $$\Phi(N)$$ and deriving $$N=\Phi(N)(N\cap M)$$, a contradiction.)

But, as there may not exist a maximal subgroup of $$N$$ containing $$N\cap M$$, the case is different when some proper subgroup of $$N$$ is not contained in a maximal subgroup of $$N$$.

Hence I'd like to ask that if $$N\unlhd G$$ and $$M$$ a maximal subgroup of $$G$$, could the case that there does not exist a maximal subgroup of $$N$$ containing $$N\cap M$$ happen? Or is there another way to prove the statement? (Otherwise, is there a counterexample that the statement does not hold when some proper subgroup of $$N$$ is not contained in a maximal subgroup of $$N$$?)

In fact, I wonder that if a group satisfies the condition that every proper subgroup is contained in a maximal subgroup, could it be possible that the condition does not apply to its normal subgroup?

partial answer: Thanks to a comment I realise my answer is only sufficient in the case $$\Phi(N)$$ is finitely generated.

You don't need every proper subgroup of $$N$$ to be contained in a maximal subgroup of $$N$$ to reach $$N=\Phi(N)(N\cap M)$$.

If $$M$$ is a maximal subgroup of $$G$$ not containing $$\Phi(N)$$ then $$G=\Phi(N)M$$ so by the modular law for groups we have $$N=N\cap\Phi(N)M=\Phi(N)(M\cap N)$$

Edit:

This is a contradiction when $$\Phi(N)$$ is finitely generated because the Frattini subgroup of $$N$$ is the set of non-generators of $$N$$. That is $$N=\langle \Phi(N),M\cap N\rangle$$ implies that $$N=M\cap N$$ so $$N\le M$$. Hence $$G=\phi(N)M\le NM=M$$.

• Sorry for not making myself clear. Actually I know how to reach $N=\Phi(N)(N\cap M)$ and I'm just wondering how to proceed after that. – Wembley Inter Mar 25 at 9:53
• ah ok, will add that – Robert Chamberlain Mar 25 at 9:58
• For the edited answer, I'm a little bit uncertain about the deduction that $\langle\Phi(N),\ M\cap N\rangle=M\cap N$ when $\Phi(N)$ is the set of non-generators. As the definition of non-degenerator is exclusively about one element, I'm wondering is it plausible to extend from one element to the whole set. – Wembley Inter Mar 25 at 12:08
• For instance, since there exists group $N$ without maximal subgroup, $e.g.$, the Prüfer group, whose Frattini subgroup is just itself, then $N$ is generated by $\Phi(N)$ and any of its subgroup (even the trivial subgroup). Subsequently, by the same deduction, will get $N$ equals any of its subgroup? That seems not so true... – Wembley Inter Mar 25 at 12:09
• It is true when $G$ has "enough" maximal subgroups, that is,when any proper subgroup is contained in a maximal subgroup; this is the case for finite groups for instance, or finitely generated groups. – Max Mar 25 at 12:32

I've posted the question on MathOverflow and get a marvelous counterexample constructed by Ycor.