Proof about maximal subgroups of a group 
Let $(G, \cdot)$ be  group, and $M<G$ normal, we say that $M$ is a normal maximal subgroup of $G$ if don't exist any subgroup $H$ normal of $G$ such that $ M \subset H \subset G$.


Theorem.
Let $(G, \cdot)$ be a group, and $M$ subgroup of $G$, then $M$ is a normal maximal  subgroup if and only if $G/M$ is a simple group.

Plese give my feedback about my approach, posted below.
 A: 
Let $(G, \cdot)$ be  group, and $M<G$ normal, we say that $M$ is a normal maximal subgroup of $G$ if don't exist any subgroup $H$ normal of $G$ such that $ M \subset H \subset G$.


Theorem.
Let $(G, \cdot)$ be a group, and $M$ subgroup of $G$, then $M$ is a normal maximal  subgroup if and only if $G/M$ is a simple group.

Proof
Assume $(G, \cdot)$  a group and $M$ normal maximal subgroup of $G$, we should prove that $G/M$ is a Simple group.
Suppose $G/M$ not are a simple group, then there exists $H/M<G/M$ normal non trivial (by our assumption), such that $H<G$ normal non trivial such that $H$ $"contains"$ $M$, which is a contradiction with the maximality of $M$, therefore $G/M$ is a simple group.
Conversely, suppose  $(G, \cdot)$   be a group and $G/M$ simple group, we would show that $M$ is normal maximal subgroup  of $G$.
Suppose $M$ not are normal maximal subgroup of $G$, then by our assumption there $H<G$ normal non trivial such that $M \subset H \subset G$ , now I claim that $H/M<H/G$ normal, its follow by the fact that $H$ is normal subgroup
of $G$  and $H/M$ is normal subgroup of $G$ non trivial, therefore $G/M$ not are simple group which is a contradiction with our hypothesis of $G/M$ is simple group.
A: Hint:  Let $\pi:G\to G/M$ be the canonical projection.  Then any normal subgroup of $G/M$ corresponds under $\pi$ to a normal subgroup of $G$ containing $M$.
