$M$ 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

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Isn't this just a straight forward application of the correspondence theorem? – user1729 Jul 29 '11 at 7:49
Yes, as mentioned in Fredrik's comment. But in the book, ideal correspondence theorem is not revealed, so it can be understandable from easier facts, as in Brian's answer. Actually, I doubt that Fraleigh's book has no contents about the correspondence theorem. – Gobi Jul 29 '11 at 8:46
– François Muer Feb 8 '14 at 15:46

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.
Apply $\phi^{-1}$ on $N/M=G/M$. What happens? Remember the definition of maximal subgroup.
Applying $f^{-1}$ just yields that $N \subseteq f^{-1}(f(N))=f^{-1}(f(G))=G$. Am I wrong? – Gobi Jul 29 '11 at 5:49