Let $(a,z) \mapsto a*z=e^{ia}z$ be a application such that $a \in \mathbb{R}$ and $z \in S=\{ z \in \mathbb{C} : |z|=1\}$. Is this a faithful action?

Question

Let $ a, b \in \mathbb {R} $ such that $ a \neq b $ and $ x \in S $. We have that, $ a * x = e^{ia}x $ and $ b * x = e^{ib} x$. Is it possible to conclude that $ a * x \neq b * x $ is a faithful action?

I do not know if it really is a faithful action, I need help.

Answer 1

If $e^{ia} = e^{ib}$, then $e^{i(b-a)} = 1$, and so $a \equiv b \bmod 2\pi$.

Answer 2

For concreteness, we can show that a representation is not faithful by showing it has non-trivial kernel. Take $2\pi \in \mathbb{R}$. Then under the representation map, it goes to $$ 2\pi \mapsto \varphi $$ such that $\varphi(x) = e^{i2\pi}(x)$. We compute for general $x \in S$, $$ \varphi(x) \equiv e^{2\pi i}(x) = 1x = x,\ \ \text{i.e. }\ \varphi(x) = x $$ So $2\pi \mapsto \varphi$ acts the exact same, as a transformation on $S$, as the identity transformation. Thus $2\pi$ is in the kernel of the representation. As $2\pi$ is not the identity element of $\mathbb{R}$, the kernel has more than one element, so the representation is non-faithful.