Let $G$ be a finite nilpotent group and $M$ be a maximal subgroup of $G$. If $H$ is a proper non-trivial subgroup of $G$ such that $H\not\leq M$, then we can show that $H\cap M$ is a maximal subgroup of $H$. Because every maximal subgroup of $G$ is normal and using second isomorphism theorem implies that $|H:H\cap M|$ is prime.
Could we generalize this property for NON-nilpotent groups?
Any proof or counterexample will be appreciated! Thanks in advance.
According to spin's nice point let me add this assumption to the question:
"Let $H\cap M\neq 1$".