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Let $G$ be a locally compact abelian infinite group but non-compact.

In some paper, the author claims that the dual group $\widehat{G}$ contains an infinite discrete group $K$.

What do you think about this? Is it true?

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up vote 1 down vote accepted

Suppose that $G$ is $\mathbb Z$, the integers. This group is discrete and so is locally compact but not compact. It is an abelian infinite group with dual group $\hat G= \mathbb T$, the circle group. (By this I mean $$\mathbb T = \{ z \in \mathbb C \ : \ |z|=1 \}$$ with the multiplication inherited from $\mathbb C$.)

This is a compact abelian group. Any infinite subgroup $K$ of $\mathbb T$ will not be discrete in the topology inherited from $\mathbb T$. If $K$ were discrete, then every subset of $K$ would be closed. But as $K \subset \mathbb T$ is infinite, it must contain a point $x \in K$ and a sequence of distinct points $x_n \neq x$ such that $x_n \to x$. If $K_0=\{x_n\}$, then $K_0$ is not closed as it does not include $x$.

This contradicts the statement in the question so check to see if there are any additional assumptions being made in the paper you are reading.

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It seems to me that your proof is incomplete ("it must contain"...). But it is true that any non-trivial discrete subgroup of the unit circle is a finite cyclic group and you arrive to the same conclusion than me. The forthcoming and accepted paper contains a serious gap... –  Zouba Nov 7 '12 at 13:51
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