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In the book "Kazhdan's Property (T)" by Bekka-de la Harpe-Valette, the following is stated on p.6 of the introduction:

"Of course, it is classical that arithmetic lattices are finitely generated, indeed finitely presented."

They cite Thm. 6.12 in "Arithmetic subgroups of algebraic groups" by Borel-Harish-Chandra for this. However, that theorem states that if $G$ is an algebraic group over $\mathbb{Q}$, then $G(\mathbb{Z})$ is finitely generated. Why is it also finitely presented?

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    $\begingroup$ A better reference is to the work of Borel and Serre who construct a compactification of the associated locally symmetric space/orbifold as a manifold/orbifold with corners. Finite presentability (and much more) follows. $\endgroup$
    – Moishe Kohan
    Commented Mar 13 at 13:07
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    $\begingroup$ Moishe provides the argument when $G$ is reductive. For the general case, one has to boil down to the latter case. Namely in this case, $U$ being the unipotent radical, this boils down to the result for $G/U$, the lattice being, up to finite index, extension with kernel $U(\mathbf{Z})$ (which is f.g. nilpotent, hence finitely presented) and quotient $(G/U)(\mathbf{Z})$. $\endgroup$
    – YCor
    Commented Mar 13 at 14:27
  • $\begingroup$ I see, thanks a lot for your replies! $\endgroup$
    – studiosus
    Commented Mar 13 at 16:52

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