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Jun 7, 2010 at 5:30 history edited Torsten Ekedahl CC BY-SA 2.5
Added a few more details.
Jun 7, 2010 at 1:14 comment added Victor Protsak Example of (2): matrices $g=\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$ and $h=\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}$ aren't even conjugate in $SL_2(\mathbb{R})$, but the cyclic subgroups they generate are conjugate by $\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$, which maps $g$ into $h^{-1}.$
Jun 7, 2010 at 0:56 comment added Victor Protsak While that surely plays a role, still more work is needed: (1) classification of integral representations of $\mathbb{Z}/p\mathbb{Z}$ does not coincide with the classification of $\mathbb{Z}[\zeta_p]$-modules, cf Curtis and Reiner, (74.3); (2) conjugation by $SL_2(\mathbb{Z})$ is more restrictive than conjugation by $GL_2(\mathbb{Z})$, which is counterbalanced by the fact that cyclic subgroups $\langle g\rangle$ and $\langle h\rangle$ may be conjugate without their generators $g$ and $h$ being conjugate.
Jun 6, 2010 at 19:17 history answered Torsten Ekedahl CC BY-SA 2.5