Let T is a hausdorff group topology and (G,T) is locally compact abelian group.If (G,T) has no open compact subgroups then can we say G has an infinite discrete cyclic subgroup?
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$\begingroup$ by discrete, I assume you mean in the subspace topology? $\endgroup$– Yemon ChoiCommented Jun 30, 2010 at 20:05
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1$\begingroup$ "Let T is a hausdorff group topology and (G,T) is locally compact abelian group." There is some redundancy in this definition/condition. $\endgroup$– Yemon ChoiCommented Jun 30, 2010 at 20:06
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$\begingroup$ Perhaps you should emphasize that you are not allowing G itself to be compact. $\endgroup$– S. Carnahan ♦Commented Jun 30, 2010 at 23:47
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$\begingroup$ here is the article 4shared.com/document/ih8d9nMA/… and please see theorem 3.2 $\endgroup$– user7188Commented Jul 1, 2010 at 6:23
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1 Answer
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This is a comment. I'm putting it as an answer just to draw attention to the question. (I hope it's ok.) It might be very easy. I'm very curious to know the answer. I'd state the question as follows:
Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?
EDIT OF AUG. 1, 2010
By page 110 of Weil's book [1], the answer is No in the abelian case.
A particular case of the question in the nonabelian case is this. Let G be a non-compact connected Lie group. Does G admit a discrete infinite cyclic subgroup? Again, I suspect that this subquestion is very easy (at least for experts).
[1] Weil, André, L'intégration dans les groupes topologiques et ses applications, Actual. Sci. Ind., no. 869. Hermann et Cie., Paris, 1940. 158 pp.
answered Jul 31, 2010 at 11:33
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$\begingroup$ Pierre-Yves, I think this could and should be posted as a separate question, since it is more precise and a bit more interesting than the original, overly vague question of the original poster. $\endgroup$ Commented Aug 1, 2010 at 21:43