If $N\triangleleft G$ then $\Phi(N)<\Phi(G)$.

Question

For any group $G$, the group $\Phi(G)$ is defined by the intersection of all maximal subgroups of $G$.

If $N\triangleleft G$ then $\Phi(N)<\Phi(G)$.

Proof. One known is that $\Phi(G)$ is a characteristic subgroup of $G$. Suppose $\Phi(N)\not<M$ for some maximal subgroup of $G$. Then as $\Phi(N)$ is a characteristic subgroup of $N$ and $N$ is normal in $G$, $\Phi(N)$ is normal in $G$. By the isomorphism theorem, $\Phi(N)M<G$ and by the maximality of $M$, $\Phi(N)M = G$.

I don't know what to do next. Could you help?

Answer 1

1. $N = G\cap N = \Phi(N)M\cap N = \Phi(N)(M\cap N)$.
2. $\Phi(N)\not< M\Rightarrow N\not<M$.
3. $M\cap N$ is properly contained in $N$ and any maximal subgroup of $N$ contains $\Phi(N)$.
4. $\Phi(N)(M\cap N)$ is contained in some maximal subgroup of $N$.