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I want to show that $\left\{\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} : \theta \in \mathbb{R} \right\}$ and $\left\{\begin{pmatrix} t & 0 \\ 0 & \frac{1}{t} \end{pmatrix} : t >0 \right\}$ are not conjugate in $SL(2,\mathbb{R})$. To do so, can I just say that the former has complex eigenvalues while the latter has real eigenvalues and conjugate matrices must have the same eigenvalues??

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    $\begingroup$ Even stronger: they're not isomorphic as abstract groups (since only the first has torsion) nor homeomorphic (since only the first is compact). $\endgroup$ – anon Sep 12 '17 at 1:56
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Yes, you can argue in that way. Or you could say the first subgroup is compact, but the second isn't, or the first subgroup has nontrivial elements of finite order, but the second doesn't.

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  • $\begingroup$ they're definitely both connected though, right? $\endgroup$ – okaskadasd Sep 12 '17 at 1:59
  • $\begingroup$ Yes. Although one is simply connected and the other isn't. $\endgroup$ – anon Sep 12 '17 at 2:00
  • $\begingroup$ @LordSharktheUnknown Actually can you say why the difference in compactness means they can't be conjugate? Is it just because conjugation is continuous? $\endgroup$ – okaskadasd Sep 12 '17 at 2:59
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    $\begingroup$ @okaskadasd conjugation by an element is a homeomorphism from $\text{SL}_2(\Bbb R)$ to itself. $\endgroup$ – Lord Shark the Unknown Sep 12 '17 at 3:01

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