On Normality of a subgroup of a finite group

Question

Let $G$ be a finite group such that $G=AB$, where $A$ is a maximal subgroup of $G$ and $B$ is nilpotent also $B \unlhd G$. Then why $A \cap B \unlhd G$?

Answer 1

Since $B \unlhd G$, we get $A \cap B \unlhd A$. So to prove that $A \cap B \unlhd G$, you only need to show that $A \cap B \unlhd B$. If that is proved then certainly $A \cap B \unlhd G=AB$. We show that $A \cap B$ is maximal in $B$. Since $B$ is nilpotent, this implies that $A \cap B \unlhd B$. Assume that $A \cap B \subsetneq M \subsetneq B$ is a subgroup of $B$. We can choose $M$ to be maximal, whence normal by the nilpotency of $B$. So $AM$ is a subgroup, hence, $A$ being maximal, there are two cases: $AM=A$ or $AM=G$. In the first case, $M \subseteq A$. But also $M \subseteq B$, so $M \subseteq A \cap B$, a contradiction. In the latter case, $AM=AB$, so $B=B \cap AM=$ (applying Dedekind's Rule) $=(B \cap A)M=M$, again a contradiction.