I'm looking for a example of a Lie group compact, abelian and disconnected, such that exist some elements $x$, where $x^n\neq e, \forall n\in\mathbb{N}$.
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Can you try some subset of $O(n,\mathbb{R})$? – DiegoMath Oct 01 '13 at 15:20
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O(n,$\mathbb{R}$) is a compact Lie group, it has two connected components. So, can you find some abelian subset of it? – DiegoMath Oct 01 '13 at 15:23
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Try irrational rotations in $U(1)\times (\mathbb{Z}/2\mathbb{Z})$. It forms such an example:
- It is the product of two abelian groups, hence abelian
- Topologically, it is a disjoint union of two copies of $S^1$, hence compact and disconnected
- Irrational rotations give infinite cyclic subgroups
Neal
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Is the $\mathbb Z_2$ meant to be integers-mod-2, or 2-adic integers? If the former, which seems to be suggested by the second bullet point, I'd be confused about how irrational rotations are essentially different from irrational rotations in a single circle, in the first place, and how the topology avoids being just that of $\mathbb Z$, etc. Please clarify the intent? Thanks! – paul garrett Oct 02 '13 at 00:33
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@paulgarrett $\mathbb{Z}_2$ is indeed $\mathbb{Z}/2\mathbb{Z}$. The irrational rotations aren't essentially different from irrational rotations in a single circle, but the OP asked for a disconnected group. The topology is just a disjoint union of two circles. (It's an uncountable set, so cannot be homeomorphic to $\mathbb{Z}$.) Does this help clarify? – Neal Oct 02 '13 at 00:57
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1Thanks, @Neal, and now as I reread the question, I see that it is asking for much less than what I'd (mis-) understood. Thanks again! – paul garrett Oct 02 '13 at 01:00
