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I need a reference or a short proof for the following property:

A nontrivial connected locally compact group $G$ contains an infinite abelian subgroup.

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Are you dealing with non-Lie groups? Because for a Lie group $G$, take $\exp{vt}$ for some $v\in \mathfrak{g}$. –  Neal Jun 28 '12 at 21:50
    
Yes, I deal with non-Lie groups. Thank you for the argument for Lie groups. –  Zouba Jun 29 '12 at 5:06
    
You'd better assume $G$ is nontrivial. –  Chris Eagle Jun 29 '12 at 12:15
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You also need $ G $ to be Hausdorff in order for the result to have any possibility of holding. For example, any non-trivial finite group with the indiscrete (trivial) topology is automatically connected and compact (hence locally compact). –  Haskell Curry Jan 31 '13 at 8:44

1 Answer 1

up vote 3 down vote accepted

I will assume, in addition, that $G$ is Hausdorff (otherwise the statement is clearly false). Then the statement is a direct corollary of the following

Theorem (Yamabe and Gleason). Let $G$ be a connected locally compact Hausdorff group. Then for every neighborhood $U$ of $1\in G$ there exists a compact normal subgroup $K<G$, so that $K\subset U$ and $L=G/K$ is isomorphic (as a topological group) to a (necessarily connected) Lie group.

See this post in Terry Tao's blog for the discussion of Y-G theorem. Note that, as Tao observes, there is no need in taking the subgroup $G'$ as we assume that $G$ is connected.

Now, given this, we note that, without loss of generality, $L$ is nontrivial (by taking $U$ small enough). A connected nontrivial Lie group always contains an infinite cyclic subgroup $Z$ (see Neal's comment). Now, take a generator $z\in Z$ and let $g\in G$ be its preimage under the epimorphism $G\to L$. Then the subgroup $<g>$ generated by $g$ is an infinite cyclic subgroup of $G$.

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