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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?

Wembley Inter
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3 Answers3

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

Robert Chamberlain
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  • 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 '19 at 09:53
  • ah ok, will add that – Robert Chamberlain Mar 25 '19 at 09:58
  • 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 '19 at 12:08
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    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 '19 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. – Maxime Ramzi Mar 25 '19 at 12:32
  • @WembleyInter You are correct, it is not too hard to make the same deduction when $\Phi(N)$ is finitely generated, so I will make a note of this in the answer – Robert Chamberlain Mar 25 '19 at 12:45
  • @RobertChamberlain Thanks! Still I'm a little bit curious about whether the statement is correct in the case not finitely generated and, moreover, if a group satisfies the condition that every proper subgroup is contained in a maximal subgroup, whether it is possible that the condition does not apply to its normal subgroup. Hope for your answer~ – Wembley Inter Mar 25 '19 at 13:01
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I've posted the question on MathOverflow and get a marvelous counterexample constructed by Ycor.

Wembley Inter
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As for whether $N$ needs to have the property that every proper subgroup is contained in some maximal subgroup of $N$. Wouldn't Klein-4 group $V_4=\langle a,b|a^2=b^2=1, ab=ba\rangle$ be a counterexample?

Take any subgroup $\langle a\rangle$, which is normal in $V_4$, and $\Phi(\langle a\rangle)=\langle a\rangle$ but $\Phi(V_4)=1$?

Adam Yin
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  • Actually we have $\Phi(\langle a\rangle) = 1$, as $1$ is a maximal subgroup of $\langle a \rangle$ – Lukas Heger Mar 06 '22 at 21:41
  • Yes, I was trying to invoke the fact that $\Phi\langle a\rangle=\langle a\rangle$ as I thought $\langle a\rangle$ did not have maximal subgroups. Though if $1$ is maximal, then wouldn't every group has a maximal subgroup? When will the clause for no maximal subgroups ever apply? – Adam Yin Mar 06 '22 at 22:10
  • $1$ is maximal in a group iff that group is cyclic of prime order, not in every group. As for an example with no maximal subgroups, $\Bbb Q$ has no maximal subgroups. – Lukas Heger Mar 06 '22 at 22:46
  • Thanks, I got it now. For some reason, initially I thought that 1 cannot be maximal by definition of maximal subgroup but realized there was no such clause in its definition. – Adam Yin Mar 07 '22 at 17:18