Assume $G$ is a finite group such that every maximal subgroup of $G$ is normal in $G$ and for $H \leq G$ we have that $HG’=G$,then show that $H=G$.
Where $G^\prime$ is the commutator subgroup (AKA derived subgroup) of $G$ defined as $$G’=\langle \left\{\left[a,b\right]:a,b\in\ G\right\}\rangle.$$ I know that Nilpotent iff every maximal subgroup is normal,but I’m not sure if that is helpful. It would be nice if someone helps me,thanks in advance.
If $M$ is maximal and hence normal, then $[G:M]=p$ is prime (since $G/M$ has no proper nontrivial subgroups and so is cyclic of prime order), and hence $G/M$ is abelian. Therefore, $G'\subseteq M$. Thus, $G'$ is contained in each maximal subgroup.
If $H\neq G$, then let $M$ be a maximal subgroup containing $H$. Then $HG'\leq MG' = M\neq G$. Thus, if $HG'=G$, then $H=G$.