# Dual groups in Fourier analysis

For finite abelian groups and the additive group of real numbers, the dual group is isomorphic to itself.

But for the unit circle in the complex plane, the dual group is isomorphic to $\mathbb{Z}$.

I know why it should, but I don't see where does this difference come from. Is there any topological concern(like the fundamental group, etc..) in this regard or any other relative algebraic reasons like the degree of a map?

If there is any relative notion, how does it affect to the Dual group of an abelian group?

One thing that's worth noting is that the dual of a compact group is a discrete group, and vice versa. So the dual of, for example, $S^1$ with the usual topology could not be $S^1$ because it's not discrete. Similarly for $\Bbb Z$. In this way you can see finite groups are dual to finite groups thanks to the fact that compact + discrete implies finite--even aside from the canonical embedding of $G$ into $\widehat{\widehat{G}}$. For $\Bbb R$ you get lucky because of the specific nature of continuous homomorphisms into $S^1$ from $\Bbb R$ and the fact that the only closed subgroups of $\Bbb R$ are $\Bbb R$ and $\alpha\Bbb Z$ for some $\alpha\in\Bbb R$, in particular the kernel of the homomorphism into $S^1$ is such a closed subgroup--being the inverse image of the closed set $\{1\}\subseteq S^1$.