If a maximal subgroup is normal, it has prime index

I'm having trouble with an exercise in "Introduce to the theory of group" book. This is problem:

Let $M$ be a maximal subgroup of $G$. Prove that if $M$ is a normal subgroup of $G$, then $[G: M]$ is finite and equal to a prime.

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Hint: there's one main thing we do with normal subgroups, which is to quotient by them. – Qiaochu Yuan Oct 14 '12 at 1:52

Hints:

1) If $\,N\triangleleft G\,$ , then there is a $\,1-1\,$ correspondence between subgroups of the quotient $\,G/N\,$ and subgroups of $\,G\,$ containing $\,N\,$ (this correspondence also respects index and normality, btw).

2) A group has no nontrivial subgroups (i.e., its only subgroups are the group itself and the trivial group $\,\{1\}\,$) iff it is finite of prime order or the trivial group itself.

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Point (1) is sometimes called the correspondence theorem or the lattice theorem. – user1729 Oct 16 '12 at 9:40

Notation: We denote the normal subgroup by $N$ instead.

By the Correspondence Theorem, there exists a bijection from the set of all subgroups $H$ such that $N\subseteq H\subseteq G$ onto the set of all subgroups of $G/N$. Since the only such subgroups are $H=N$ and $H=G$, $G/N$ has only two subgroups, namely $N/N$ and $G/N$.

Let $xN$ be a nontrivial element in $G/N$. $\langle xN\rangle$ is a nontrivial subgroup of $G/N$, thus $\langle xN\rangle=G/N$. This means $G/N$ is cyclic. If $|G/N|$ is infinite, then $G/N\cong\mathbb{Z}$ which is a contradiction as $\mathbb{Z}$ has infinite subgroups of the form $n\mathbb{Z}$. Therefore $[G:N]=|G/N|$ is finite.

Thus $G/N\cong\mathbb{Z}/n\mathbb{Z}$ for some integer $n$. By the Correspondence Theorem, the subgroups of $\mathbb{Z}/n\mathbb{Z}$ are $m\mathbb{Z}/n\mathbb{Z}$ where $n\mathbb{Z}\subseteq m\mathbb{Z}\subseteq\mathbb{Z}$. This means $m\mid n$.

Since $G/N$ has two subgoups, this means $n$ has exactly 2 divisors, so $n$ is a prime. Thus $[G:N]=|\mathbb{Z}/n\mathbb{Z}|=n$ is a prime.

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