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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?

flag	asked Jun 30 2010 at 18:46 unknown (google) 15●1

by discrete, I assume you mean in the subspace topology? – Yemon Choi Jun 30 2010 at 20:05

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"Let T is a hausdorff group topology and (G,T) is locally compact abelian group." There is some redundancy in this definition/condition. – Yemon Choi Jun 30 2010 at 20:06

Perhaps you should emphasize that you are not allowing G itself to be compact. – S. Carnahan♦ Jun 30 2010 at 23:47

here is the article 4shared.com/document/ih8d9nMA/… and please see theorem 3.2 – unknown (google) Jul 1 2010 at 6:23

Pierre-Yves Gaillard · Answer 1 · 2010-08-02 05:59:07Z UTC

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.

Discrete cyclic subgroup.

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