Why $ K \cap H $ is a maximal subgroup of $H $? Suppose $ G $ is a finite group and $ H \unlhd G $. Suppose that $ H \cap M $ is either $ H $ or a maximal subgroup of $ H $ for any maximal subgroup $ M $ of $ G $. Let $ N $ be a minimal normal subgroup of $ G $.  Suppose $ M/N $ is a maximal subgroup of $ G/N $. Then $ M/N \cap HN/N = ( M \cap H )N/N $ equal $ HN/N $ or a maximal subgroup of $ HN/N $. Suppose $ K/N $ is a maximal subgroup of $ HN/N $, then $ K/N = K/N \cap HN/N = (K \cap H)N/N $. Why $ K \cap H $ is a maximal subgroup of $H $? 

If $N$ is contained in $H$, then this is trivial, since then $K$ is contained in $HN=H$, and so by assumption $K=K\cap H$ is maximal in $H$. 
Otherwise, by the minimality of $N$, we have $H\cap N$ is trivial. Suppose there exists $M$ with
$$ K\cap H \le M \le H$$
Then we also have 
$$ K=(K\cap H)N \le MN \le HN $$
Since all the groups on the left have trivial intersection with $N$, and $K$ is maximal, we have $|M|=|K\cap H|$ or $|M|=|H|$. Thus $M$ is $K\cap H$ or $H$. Thus $K\cap H$ is maximal. 
