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YCor
YCor
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Does anybody know what is the biggest $r$ such that $\mathbb{Z}^r$ is isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$?

It cannot be bigger that the virtual cohomological dimension of $\mathrm{SL}_n(\mathbb{Z})$, which is $\frac{n(n-1)}{2}$, since the cohomological dimension respects inclusions. But I suspect it must be smaller.

Does anybody know what is the biggest $r$ such that $\mathbb{Z}^r$ is a subgroup of $\mathrm{SL}_n(\mathbb{Z})$?

It cannot be bigger that the virtual cohomological dimension of $\mathrm{SL}_n(\mathbb{Z})$, which is $\frac{n(n-1)}{2}$, since the cohomological dimension respects inclusions. But I suspect it must be smaller.

Does anybody know what is the biggest $r$ such that $\mathbb{Z}^r$ is isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$?

It cannot be bigger that the virtual cohomological dimension of $\mathrm{SL}_n(\mathbb{Z})$, which is $\frac{n(n-1)}{2}$, since the cohomological dimension respects inclusions. But I suspect it must be smaller.

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Luis Jorge
Luis Jorge
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Free abelian subgroups of $\mathrm{SL}_n(\mathbb{Z})$

Does anybody know what is the biggest $r$ such that $\mathbb{Z}^r$ is a subgroup of $\mathrm{SL}_n(\mathbb{Z})$?

It cannot be bigger that the virtual cohomological dimension of $\mathrm{SL}_n(\mathbb{Z})$, which is $\frac{n(n-1)}{2}$, since the cohomological dimension respects inclusions. But I suspect it must be smaller.