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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!

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Indeed the result is false.

Consider the affine group $G=\mathbf{Q}^*\ltimes\mathbf{Q}$ and $N$ the normal subgroup $\mathbf{Q}$.

Since $N$ has no maximal proper subgroup $\Phi(N)=N$.

Since $\mathbf{Q}^*$ is a maximal proper subgroup of $G$ and since the intersection of its conjugates is trivial, we have $\Phi(G)=\{1\}$. So $\Phi(N)\nsubseteq \Phi(G)$.

The problem is exactly what you're pointing out. The result however holds whenever $N$ has the property that every proper subgroup of $N$ is contained in a maximal proper subgroup of $N$.

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