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

If $V$ is an $n$-dimensional real vector space, a lattice in $V$ is a subgroup of the form $\Gamma=\mathbb{Z}v_1+\dots+\mathbb{Z}v_m$ where $v_1,\dots,v_m\in V$ are are $\mathbb{R}$-linearly independent. The lattice is complete if $m=n$.

Theorem: A subgroup $\Gamma\subset V$ is a lattice iff it is discrete.

This is proven in Neukirch's Algebraic Number Theory but there is a step I don't understand. Here's his proof:

enter image description here

In the last step, why does $q\Gamma\subset \Gamma_0$?

UPDATE: Also, in the line just before that, why does that end the proof that $(\Gamma:\Gamma_0)$ is finite? I see it proves that there are finite $\mu_i$, but why finite $\gamma_i$?

share|cite|improve this question
up vote 4 down vote accepted

The additive group $\Gamma/\Gamma_0$ has order $q$, so for every $\gamma+\Gamma_0\in\Gamma/\Gamma_0$, $q(\gamma+\Gamma_0)=0+\Gamma_0=\Gamma_0$. But $q(\gamma+\Gamma_0)=q\gamma+\Gamma_0$, so $q\gamma+\Gamma_0=\Gamma_0$, and therefore $q\gamma\in\Gamma_0$. Since $\gamma\in\Gamma$ was arbitrary, $q\Gamma\subseteq\Gamma_0$.

share|cite|improve this answer
So this is true in general: if $G$ is a group and $H$ is a subgroup of finite index $q$, then $qH\subset G$. I thought it was something specific to the groups in the proof. Thank you! – user46225 Nov 29 '12 at 20:15
@user46225: You have it backwards: it should be $qG\subseteq H$. That’s true in general for additive groups; for a multiplicative group it would be $G^q\subseteq H$. – Brian M. Scott Nov 29 '12 at 20:17
Dear Brian, there's another step in the proof which I don't understand. Why is the set $\{\gamma_i\}$ finite? I understand why the $\{\mu_i\}$ are finite, but I don't see why it implies the $\{\gamma_i\}$ are as well. – user46225 Nov 29 '12 at 22:16
@user46225: Because the map $\gamma_i\to\mu_i$ is a bijection. Suppose that $\mu_i=\mu_j$; then $\gamma_i-\gamma_j\in\Gamma_0$, and therefore $i=j$, since the $\gamma_i$’s were all from different cosets of $\Gamma_0$. – Brian M. Scott Nov 29 '12 at 22:55

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.