Let $G$ be any subgroup of the circle $\mathbb{T}$ and endow $G$ with the discrete topology. Let $$\text{id}\colon G\to\mathbb{T}$$ be the inclusion map. Why is $$\{\text{id}^{n}:n\in\mathbb{Z}\}$$ dense in the (compact) Pontryagin dual group $\widehat{G}$?
I tried to work with nets, but I struggled with the topology on $\widehat{G}$ and it got quite messy. Is there a more elegant argument?
This is essentially immediate from Pontryagin duality. Let $H$ be the closure of the subgroup of $\widehat{G}$ generated by id. If $H$ is not all of $\widehat{G}$, then there is a nontrivial character $\widehat{G}\to\mathbb{T}$ that is trivial on $H$. But every character on $\widehat{G}$ is evaluation at some element of $G$, so this gives a nontrivial element $g\in G$ such that $h(g)=1$ for all $h\in H$. In particular, taking $h=\mathrm{id}$, this says $g=1$, which is a contradiction.
More generally, a similar argument shows that if $f:G\to T$ is an injective homomorphism of locally compact abelian groups, the dual homomorphism $\widehat{T}\to\widehat{G}$ has dense image. (In your case $T=\mathbb{T}$ and $f=\mathrm{id}$, and its dual is just the inclusion $\mathbb{Z}\to\widehat{G}$ of the subgroup generated by id.)