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**Statement:**

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

**Proof:**

Let $M$ be a maximal normal subgroup of $G$. Consider the canonical homomorphism $\gamma : G \rightarrow G/M$ given by Theorem 14.9. Now $\gamma ^{-1}$ of any nontrivial proper normal subgroup of $G/M$ is a proper normal subgroup of $G$ properly containing $M$. But $M$ is maximal, so this can not happen. Thus $G/M$ is simple.

Conversely, Theorem 15.16 shows that if $N$ is a normal subgroup of $G$ properly containing $M$, then $\gamma [N]$ is normal in $G/M$. If also $N \neq G$, then

$\gamma [N] \neq G/M$ and $\gamma [N] \neq \{M\}$.

Thus, if $G/M$ is simple so that no such $\gamma [N]$ can exist, no such $N$ can exist, and $M$ is maximal.