This is an exercise form Katznelson's book on Harmonic Analysis, so I want to solve it using his hint.
T* here denotes the multiplicative group of units of complex numbers of unit norm. That is to say T* = {$z \in \mathbb{C}$ | $|z|=1$ }
In a previous exercise, I was successful in showing that any infinite subgroup G of $A = \mathbb{R}/(2\pi\mathbb{Z})$ must be dense in $A$.
The hint given is to "determine the mapping on 'small' rational multiples of $2\pi$ and use [the exercise stated above]."
First, I observed that the homomorphism must send the identity to the identity hence 0 must get sent to 1.
So I have attempted this problem in various ways. I'm just not sure how to define G. I considered defining it as the group of all rational multiples of $2\pi$ (mod $2\pi$). But this is not getting me anywhere: small rational multiples will converge to 0 hence their images must converge to 1 by continuity. So what...?
I can also guess (intuitively, without proof) that the family of homomorphisms which would work must be of the form $x \mapsto e^{irx}$ for some $r \in \mathbb{R} $.
So in some sense, we need to show that an arbitrary continuous homomorphism should be of that form.
I have also observed that if we write down some $G$, and $G$ is dense in $A$ then by continuity the homomorphism is determined by its behavior on G.
But really, how do you solve this problem? Thank you all for your suggestions!