On page 199 of Dummit and Foote's Abstract Algebra (Here $\Phi(G)$ is the Frattini subgroup of a group $G$, not necessarily finite):
If $N\unlhd G$, then $\Phi(N)\subseteq\Phi(G)$.
First, When every proper subgroup of $N$ is contained in a maximal subgroup of $N$, I know how to prove the statement. (By taking $M$ as a maximal subgroup of $G$ that fails to contain $\Phi(N)$, deriving $N=\Phi(N)(N\cap M)$ and taking the maximal subgroup $H$ of $N$ containing $N\cap M$, then $\Phi(N)\subset H$, a contradiction.)
But in the general case, as there may not exist a maximal subgroup of $N$ containing $N\cap M$, the case is different. The process of deriving $N=\Phi(N)(N\cap M)$ is the same, but I find no way to proceed after that.
Hence I have several questions:
(1) If the statement still holds in the case when not all proper subgroup of $N$ is contained in some maximal subgroup of $N$? Or does there exist an counterexample?
(2) Moreover, I'm wondering that if a group $G$ 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 $N$?
I've posted the question on StackExchange, and just get a partial answer for the question(1). (Actually, by the definition of Frattini subgroup being the set of all non-generators, the statement is proved in the case of $\phi(N)$ being finitely generated).
Hence I hope for an answer for both questions! Thanks in advance!