If $N\trianglelefteq G$, then $\phi(N) \leq \phi(G)$, where $\phi(N)$ is the Frattini subgroup of $N$. I was thinking somehow to use normality of N as follows
Since N is normal, then $G/N$ will be a group, so we can consider the 
natural map
$pi \ :\ G \rightarrow G/N$, where $g \mapsto gN$, $ker(\pi) = N$.
So It is here enough to prove that a maximal subgroup M of G will contain 
$ker(\pi)$, then since $\phi(G)$ is intersection all such maximal subgroup M,  so we will get that $\phi(N) \leq \phi(G)$, but I got stuck in this specific step.
 A: The statement has been discussed here:
For Frattini subgroup $\Phi(G)$, if $N \vartriangleleft G$, then $\Phi(N) \leq \Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $\Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $\Phi(N)$ is a normal subgroup of G.
We need to show that $\Phi(N)$ is contained in $\Phi(G)$. For this, it suffices to show that $\Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $\Phi(N)$. Then, the subgroup generated by $M$ and $\Phi(N)$ is equal to $G$. Since $\Phi(N)$ is normal in $G$, we have $M\Phi(N) = G$. From that, it follows that 
$\Phi(N)(M \cap N) = N$.
Since $M$ does not contain $\Phi(N)$, $M$ does not contain $N$ either and hence 
$M \cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M \cap N$ and $\Phi(N)$. This contradicts the fact that their product is $N$.
