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