TITLE: Discrete subgroups are lattices
QUESTION [11 upvotes]: 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 $\gamma_i$?

REPLY [9 votes]: 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, $q\Gamma\subseteq\Gamma_0$.