Frattini subgroup of a subgroup I am doing an exercise for Frattini subgroups:
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Show that there exists a subgroup $K$ of $G$ with $G=KN$ and $K \cap N \leq \Phi(K)$.
I have no idea what theorem/result would give the existence of such $K$? This $K$ looks similar to a complement to $N$ but I don't think it is indeed the complement. Or maybe I have to construct such a group?
Any hints/comments are appreciated! Thanks!
 A: Consider the set of supplements to $N$ in $G$. That is, let 
$$\mathcal{X} = \{H \leq G : HN = G\}.$$
Clearly $\mathcal{X}$ is non-empty because $G \in \mathcal{X}$. Now let $K$ be a minimal element of $\mathcal{X}$, in the sense that $K$ does not contain properly any other element from $\mathcal{X}$. By definition of $\mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K \cap N \leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K \cap N$, thus $K \cap N \leq \Phi(K)$.
Suppose for a contradiction that $K \cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K \cap N$ is normal in $K$. So $K \cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K \cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K \cap N)M$. Now notice that
$$G = KN = (K \cap N)MN = (K \cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K \cap N$ is contained in every maximal subgroup of $K$ and so $K \cap N \leq \Phi(K)$, as wanted.
