# Frattini subgroup is normal-monotone

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!

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 $$N$$.