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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
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1 Answer 1

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
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