$M$ is a maximal normal subgroup iff $G/M$ is simple $M$ is a maximal normal subgroup of G if and only if $G/M$ is simple.


I have a problem in "if" part. To prove ($\Leftarrow$) direction, assume that $N$ is a normal subgroup of $G$ properly containing $M$. Let $f:G\to G/M$ be the surjective canonical map. Then $f(N)=N/M$ is a nontrivial(since $M \subsetneq N$) normal subgroup of $G/M$. Since $G/M$ is simple, $N/M=G/M$. But how can I conclude that $N=G$? Isn't it be possible that a proper subgroup of $G$ may have the same image under factoring modulo $M$?

Reference: Fraleigh p. 150 Theorem 15.18 in A First 
Course in Abstract Algebra 7th ed
Suppose that $N/M = G/M$ and there is some $g \in G \setminus N$; then there is an $n \in N$ such that $gM = nM$. In particular, $gm = n$ for some $m \in M$, whence $g = nm^{-1} \in N$, which is a contradiction.
$(G/M)/(N/M)$ is isomorphic to $G/N$. Since $G/M$ is equal to $N/M$. Then $G/N$ is isomorphic to the trivial subgroup. That means $G/N$ is isomorphic to the trivial subgroup too. Therefore $G=N$.