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 $$N=N\cap\Phi(N)M=\Phi(N)(M\cap N)$$


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 $G=\phi(N)M\le NM=M$.

  • $\begingroup$ 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. $\endgroup$ – Wembley Inter Mar 25 at 9:53
  • $\begingroup$ ah ok, will add that $\endgroup$ – Robert Chamberlain Mar 25 at 9:58
  • $\begingroup$ 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. $\endgroup$ – Wembley Inter Mar 25 at 12:08
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    $\begingroup$ 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... $\endgroup$ – Wembley Inter Mar 25 at 12:09
  • $\begingroup$ 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. $\endgroup$ – Max Mar 25 at 12:32

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


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