Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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


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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.