Define a homomorphism $f_t$ from $\mathbb{Z}$ to $U(1)$ by $$f_t: n \rightarrow \exp {(i2\pi n t)},\,\,\,\,\,\, n\in \mathbb{Z}$$ where $t\in[0,1)$. Obviously if $t=1/m$ with $m\in\mathbb{Z}^+$, the quotient group $U(1)/\mathbb{Z}$ is isomorphic to $U(1)$. I think this can be generalized to the case when $t$ is a rational number between $0$ and $1$. My question is if $t$ is an irrational number, what is the quotient group $U(1)/\mathbb{Z}$?
Any comment, proof, or references will be appreciated!
NOTE: I made a stupid mistake here. For $t\in \mathbb{Q}$, the image of the map $f_t$ is not $\mathbb{Z}$. But I think for irrational $t$, the image will be $\mathbb{Z}$. So I think $\mathbb{Z}$ is a subgroup of $U(1)$. Please correct me if I am wrong.
The map $e^{i2\pi \theta}\mapsto e^{i2\pi \theta/t}$ is a homomorphism from $U(1)$ onto $U(1)$ that has kernel given by the image of the map you specified above. By the first isomorphism theorem, we see that the quotient group is $U(1)$.