Some properties of the Frattini subgroup

Question

I'm stuck on a question about some properties of the Frattini subgroup:

Let $K$ be a normal subgroup of a finite group $G$. Then

1. If $K$ is a proper subgroup of $G$, then $\dfrac{\Phi(G)K}{K} \leqslant \Phi\left(\dfrac{G}{K}\right)$.
2. If $K\leqslant \Phi(G)$, then $\dfrac{\Phi(G)}{K} = \Phi\left(\dfrac{G}{K}\right)$.

I don't know to prove it. Assuming item 1 proved, here's what I've done so far for item 2:

Well, since $G$ is finite we can't have $\Phi(G) = G$ and so $K \leqslant \Phi(G) < G$, that is, $K$ is a proper subgroup. Now we apply item 1 and obtain $$ \dfrac{\Phi(G)}{K} \overset{K\leqslant \Phi(G)}{=} \dfrac{\Phi(G)K}{K} \leqslant \dfrac{\Phi(G)}{K}. $$ So we have one side of the inclusion. For the other inclusion, let $gK\in \Phi(G/K)\leqslant G/K$. Then $gK$ belongs to every maximal subgroup $M/K$ of $G/K$. Then for every maximal subgroup $M$ of $G$ we have an element $h_M\in M$ such that $gK = h_MK$... and I don't know where to go from here.

Here are some other properties that may be helpful:

Let $H$ be a subgroup of $G$, then

1. If $K\leqslant\Phi(H)$ then $K\leqslant \Phi(G)$.
2. $\Phi(K)\leqslant \Phi(G)$.

Answer 1

Hmm, I think I solved it.

Proof of item 2:

Continuing my argument, fix an arbitrary maximal subgroup $M\leqslant G$. From $gK = h_MK$ we have $g = h_Mk$ for some $k\in K$. But by hypothesis $K\leqslant \Phi(G) \leqslant M$, which yields $k\in M$ and so $g$ is in $M$. This holds for all $M\lessdot G$. Therefore $g\in \Phi(G)$ and it follows $gK\in \Phi(G)/K$.

Also, I figured it out that item 1 can be done by definition.

Let $x = gK \in \Phi(G)K/K$, where $g\in \Phi(G)$. Then $g\in M$ for all $M\lessdot G$. By Correspondence Theorem (or 4th Iso. Theo. as Chris noted), we have $gK \in M/K$ for all $M\lessdot G$, which covers all maximal subgroups of $G/K$. Thus, $gK \in \Phi(G/K)$.

The problem is: I didn't use the fact that $K$ is a proper subgroup of $G$.