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3 questions
6
votes
1
answer
392
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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 isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$?
It cannot be bigger that the virtual cohomological dimension of $\...
4
votes
2
answers
207
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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 ...
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 ...