# discrete subgroup of locally compact abelian group

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 $K$.

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 $x$.