# existence of infinite abelian subgroup in infinite locally compact groups

1) Let $G$ be an infinite locally compact group.

Does there exist an infinite abelian locally compact subgroup of $G$?

Rem: I know that there exists an infinite abelian subgroup in every infinite compact group.

2) Does there exist structure theorems for locally compact groups which describe these groups with compact groups and abelian locally compact groups?

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## 1 Answer

1. No. Consider the Tarski Monster as a discrete (hence locally compact) group. As all proper subgroups are finite, there are no infinite abelian subgroups.

2. If you're okay with ignoring discrete errors (i.e. looking at an open subgroup), then there is a sense in which all locally compact groups are "almost" Lie groups. For a precise statement, see the Gleason-Yamabe theorem which can be found on page 22 of http://terrytao.files.wordpress.com/2012/03/hilbert-book.pdf

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Thank you. Ok, the Tarski monster is a counter-example. However, does there exist a more concrete and simple counter-example? –  Zouba Jun 13 '12 at 16:59
I'm not 100% sure, but I would guess that any closed subgroup of a locally compact group is locally compact. If this is true, then requiring the abelian subgroup to be locally compact isn't necessary, since you could always take the closure. Thus, you might as well assume the group is discrete. Then, any counter-example would have to be an infinite non-abelian group such that, at the very least, every element has finite order. I can't really think of any simple examples that don't obviously contain a big abelian subgroup. –  Kevin Jun 22 '12 at 5:26
For instance, if the group is finitely generated, it would be a counterexample to Burnside's problem. It took mathematicians over 60 years to construct such counterexamples. –  Kevin Jun 22 '12 at 5:32