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6 votes
1 answer
392 views

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

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 $\...
Luis Jorge's user avatar
4 votes
2 answers
207 views

Normalizers of the principal congruence subgroups in $\mathrm{GL}(n,\mathbf Q)$

A question quite similar to this question. Let $n \geqslant 3$ and $m \geqslant 2$ be natural numbers and suppose that a matrix $A \in \mathrm{GL}(n,\mathbf Q)$ normalizes the principal congruence ...
P.H.'s user avatar
  • 43
4 votes
1 answer
218 views

Commensurator of a subgroup of matrices

Let $k$ be a totally real number field and let $\mathcal{O}_k$ denote its ring of integers. If $H$ is a subgroup of $\text{GL}(n, \mathbb{R})$ let denote with $H(k)$ and $H(\mathcal{O}_k)$ the ...
Diego95's user avatar
  • 521