TITLE: Prove that if $A \vartriangleleft G$ is abelian, then $A$ has a complement in $G$.
QUESTION [5 upvotes]: Let $G$ be a finite group. Suppose that the intersection of all of the maximal subgroups of $G$ is trivial. Prove that if $A \vartriangleleft G$ is abelian, then there exist $U \subseteq G$ such that $AU=G$ and $A \cap U=1$.
Tell me some hints.
Thanks in advanced.

REPLY [5 votes]: Let $G$ be a minimal counterexample. We can choose $A$ such that $|A|$ is minimal. Hence we can find $M \le G$ such that $G=MA$ and $M\cap A\ne 1$. We can choose $M \ne G$ because the intersection of all maximal subgroups is trivial. Let $B=M \cap A$. Clearly $B \ne A$ by our choice. Since $A$ is abelian and $G=MA$, we get $B$ is a normal subgroup in $G$, i.e. $B$ satisfies all requirement of $A$.
By minimality of $A$, there exists $M_1$ such that $G=M_1B$ and $M_1 \cap B=1$. Hence $M=M \cap G=B(M \cap M_1)$. So $G=MA=(M_1\cap M)BA=(M\cap M_1)A$. Now $(M\cap M_1 )\cap A=M_1 \cap B=1$, a contradiction.