Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.