Minimal normal subgroup in a finite group

Question

Let $G$ be a finite group and $G=HK$ such that $H<G$. How can we show that if $K$ is an abelian minimal normal subgroup of $G$, then $H$ is a maximal subgroup of $G$ and $H\cap K=1$?

Answer 1

Let $N=H \cap K$. Then $N$ is normal in $K$ since $K$ is abelian, and $N$ is normal in $H$, since $K$ is normal in $G$. But $G=HK$, so it follows that $N$ is normal in $G$. $K$ is a minimal normal subgroup of $G$, so either $N=1$ or $N=K$. If $N=K$, then $N=H \cap K=K \subseteq H$, from which the contradiction $G=HK=H$.

Now, let $M$ be a subgroup of $G$, with $H \subseteq M \subseteq G$. Put $L= M\cap K$. The same line of reasoning applies, $L$ is normal in both $M$ and $K$, and since $H \subseteq M$, we have $G=MK$ and $L$ is normal in $G$. By the minimality of $K$, it follows that $L=1$ or $L=K$. If $L=1=M \cap K$, then by Dedekind's Law $H=H(M \cap K)=M \cap HK=M$. If $L=M \cap K=K$, then $K \subseteq M$, whence $G=HK \subseteq HM=M$, so $G=M$. This proves that $H$ is a maximal subgroup.