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  "body": "\n## Definition\n\nLet $G$ be a linear algebraic group defined over $\\Q$, together with a faithful representation $G \\hookrightarrow \\mathrm{GL}_n$ defined over $\\Q$.\nLet $G(\\Z) = G(\\Q) \\cap \\mathrm{GL}_n(\\Z)$.\nA [[subgroup|0X1JXT47GA]] $\\Gamma$ of $G(\\Q)$ is an *arithmetic subgroup* of $G$ if it is commensurable with $G(\\Z)$.\nThis means that $\\Gamma \\cap G(\\Z)$ has finite [[index|0DTXBQPA2S]] both in $\\Gamma$ and in $G(\\Z)$.\nAn *arithmetic group* is a group that is an arithmetic subgroup of some such $G$.\n\n## Remarks\n\nThe class of arithmetic subgroups does not depend on the choice of the faithful representation.\n\nExample: the [[special linear group|0V9PMN0RBH]] $\\mathrm{SL}_n$ in its standard representation has $G(\\Z) = \\mathrm{SL}_n(\\Z)$, which is arithmetic.\nFor an integer $N \\ge 1$, the [[congruence subgroup|1H6R856P3C]] $\\Gamma(N)$, the kernel of reduction modulo $N$ from $\\mathrm{SL}_n(\\Z)$ to $\\mathrm{SL}_n(\\Z/N\\Z)$, is arithmetic too.\nIndeed, $\\Gamma(N) \\cap \\mathrm{SL}_n(\\Z) = \\Gamma(N)$, and its index in $\\mathrm{SL}_n(\\Z)$ is at most the order of the finite group $\\mathrm{SL}_n(\\Z/N\\Z)$.\n\nNon-example: $\\Gamma = \\mathrm{SL}_2(\\Z[1/2])$ inside $\\mathrm{SL}_2(\\Q)$.\nHere $\\Gamma \\cap \\mathrm{SL}_2(\\Z) = \\mathrm{SL}_2(\\Z)$.\nThe 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 $k \\ne m$.\nSo $\\mathrm{SL}_2(\\Z)$ has infinite index in $\\Gamma$, and $\\Gamma$ is not arithmetic in $\\mathrm{SL}_2$.\n\nOver 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 $\\Q$.\n",
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