# Discrete subgroups are lattices

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:

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 $\gamma_i$?

-
add comment

## 1 Answer

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, $q\Gamma\subseteq\Gamma_0$.

-
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
add comment