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I was thinking somehow to use normality of N as follows

Since N is normal, then $G/N$ will be a group, so we can consider the natural map

$pi \ :\ G \rightarrow G/N$, where $g \mapsto gN$, $ker(\pi) = N$.

So It is here enough to prove that a maximal subgroup M of G will contain $ker(\pi)$, then since $\phi(G)$ is intersection all such maximal subgroup M, so we will get that $\phi(N) \leq \phi(G)$, but I got stuck in this specific step.

Dietrich Burde
  • 130,978
  • I can't see why every maximal subgroup of $G$ must contain $ker(\pi )$. And I think the claim is not true. Since evry maximal subgroup of a p-group $P$ is of index p and is normal in $P$, if I take $G=P$, $N=Q$ as such maximal subgroup of my p-group, obviously, another maximal subgroup of $P$ doesn't contain $Q$. – Display name Sep 24 '21 at 01:28

2 Answers2

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The statement has been discussed here:

For Frattini subgroup $\Phi(G)$, if $N \vartriangleleft G$, then $\Phi(N) \leq \Phi(G)$. : Counter-example

We have the following result (for finite groups it coincides with the above statement):

Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.

Proof: The Frattini subgroup $\Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $\Phi(N)$ is a normal subgroup of G. We need to show that $\Phi(N)$ is contained in $\Phi(G)$. For this, it suffices to show that $\Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction. Suppose $M$ is a maximal subgroup of $G$ not containing $\Phi(N)$. Then, the subgroup generated by $M$ and $\Phi(N)$ is equal to $G$. Since $\Phi(N)$ is normal in $G$, we have $M\Phi(N) = G$. From that, it follows that $\Phi(N)(M \cap N) = N$. Since $M$ does not contain $\Phi(N)$, $M$ does not contain $N$ either and hence $M \cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M \cap N$ and $\Phi(N)$. This contradicts the fact that their product is $N$.

Dietrich Burde
  • 130,978
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Lemma: Let $N$, $K$, and $L$ be subgroups of $G$ such that $N$ is normal in $G$ and $N\leq K$. Then $N\left(K\cap L\right)=K\cap NL$

Proof of Lemma: The forward inclusion is obvious so consider $x\in K\cap NL$ so $x=k=n\ell$ for some $k\in K$, $n\in N$, and $\ell\in L$. Then $\ell =n^{-1}k=k^{\prime}\in K$ so $x\in N\left(K\cap L\right)$. Thus the lemma.

Proof: The result is trivial if $\Phi\left(N\right)$ is contained in every maximal subgroup of $G$ so suppose not. Then $N=N\cap G=N\cap\Phi\left(N\right)M$ for some maximal subgroup $M$ of $G$. The lemma implies $N=\Phi\left(N\right)\left(N\cap M\right)$ and observe, by the Second Isomorphism Theorem (3.3.18), $\Phi\left(N\right)\left(N\cap M\right)/\Phi\left(N\right)\cong\left(N\cap M\right)/\left(\Phi\left(N\right)\cap M\right)$ so \begin{align*} N/\Phi\left(N\right) &\cong\left(N\cap M\right)/\left(\Phi\left(N\right)\cap M\right) \end{align*} The latter isomorphism is contradictorially true for $N\leq M$. The right side of the supposed isomorphism is a proper subgroup of $N/\Phi\left(N\right)$ for any other case. That is, in case $\Phi\left(N\right)<N\cap M<N$, $\left(N\cap M\right)/\Phi\left(N\right)<N/\Phi\left(N\right)$. In case $1\leq N\cap M\leq\Phi\left(N\right)$, $\Phi\left(N\right)< N/\Phi\left(N\right)$. It must be that $N\leq M$. Thus the result.

sotweed
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