Arithmetic group

Remarks

The class of arithmetic subgroups does not depend on the choice of the faithful representation.

Example: the special linear group $\mathrm{SL}_n$ in its standard representation has $G(\Z) = \mathrm{SL}_n(\Z)$, which is arithmetic. For an integer $N \ge 1$, the congruence subgroup $\Gamma(N)$, the kernel of reduction modulo $N$ from $\mathrm{SL}_n(\Z)$ to $\mathrm{SL}_n(\Z/N\Z)$, is arithmetic too. Indeed, $\Gamma(N) \cap \mathrm{SL}_n(\Z) = \Gamma(N)$, and its index in $\mathrm{SL}_n(\Z)$ is at most the order of the finite group $\mathrm{SL}_n(\Z/N\Z)$.

Non-example: $\Gamma = \mathrm{SL}_2(\Z[1/2])$ inside $\mathrm{SL}_2(\Q)$. Here $\Gamma \cap \mathrm{SL}_2(\Z) = \mathrm{SL}_2(\Z)$. The matrices $d_k = \mathrm{diag}(2^k, 2^{-k})$, $k \ge 0$, lie in $\Gamma$, and $d_k^{-1} d_m \notin \mathrm{SL}_2(\Z)$ for $k \ne m$. So $\mathrm{SL}_2(\Z)$ has infinite index in $\Gamma$, and $\Gamma$ is not arithmetic in $\mathrm{SL}_2$.

Over a number field $K$ with ring of integers $\mathcal{O}_K$, one uses $G(\mathcal{O}_K)$ in place of $G(\Z)$; restriction of scalars reduces this case to the one over $\Q$.