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](https://math.stackexchange.com/questions/3161390/), 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!



Frattini subgroup is normal-monotone
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 $N$.


