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Let $G=\mathrm{GL}_3(\mathbb{Q})$. Now, consider all subgroups in $G$ of the form $\mathbb{Z}\times\mathbb{Z}$ consisting only of unipotent elements (elements whose eigenvalues are all $1$). How many $G$-conjugacy classes of such subgroups are there? Does each conjugacy class have a representative that can be generated by commuting elements $A,B\in\mathrm{GL}_3(\mathbb{Z})$?

So far, I've found three conjugacy classes of such subgroups, with representatives given as follows:

$$\left\langle\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix},\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle\\[.4in]\left\langle\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix},\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle\\[.4in]\left\langle\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix},\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle\\$$

Is this a complete list, or are there more? It would be nice to find a solution that can be generalized to $\mathrm{GL}_n(\mathbb{Q})$ for larger $n$, although I'd even be happy with something that works for $n=4$.

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  • $\begingroup$ What do you mean by "subgroups in $G$ of the form $\mathbb{Z}\times\mathbb{Z}$ " ? How does that abelian free group relates "naturally" (?) to the group $\;G\;$ ? Or you meant all free abelian-of-rank-two subgroups? $\endgroup$ – DonAntonio Jan 11 '14 at 11:46
  • $\begingroup$ I don't feel like trying to write out proofs, but after a few calculations, I think you can always conjugate the subgroup into ${\rm GL}_3({\mathbb Z})$. But I am not convinced that there are only three classes. Suppose you change the top right entry of the second generator of your third group from $1$ to some different $t \in {\mathbb Q}$. Does that give you a conjugate subgroup? $\endgroup$ – Derek Holt Jan 11 '14 at 16:27
  • $\begingroup$ @DonAntonio: You're right, I meant what you say (all free abelian-of-rank-two subgroups). I guess I'm in the practice of not distinguishing isomorphic groups, so perhaps another way to say what I meant was "copies of $\mathbb{Z}\times\mathbb{Z}$ inside of $G$." $\endgroup$ – Jared Jan 11 '14 at 19:40

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