TITLE: Group isomorphism: $\mathbb{R}/\mathbb{Z}\cong S^1$
QUESTION [6 upvotes]: Let $\mathbb R$, $\mathbb{Z}$ be the groups of real numbers and integers respectively under addition, and $S^1$ denote the group of complex with modulus $1$ under multiplication.
Then show that $\mathbb{R}/\mathbb Z\cong S^1$.
My idea is to build a homomorphism with kernel $\mathbb Z$. I tried mapping $x\to \left(\{x\},\sqrt{1-\{x\}^2}\right)$, where $\{\cdot \}$ is the fractional part.
But its ineffective because my construction only maps to the complex number in the first quadrant.

REPLY [7 votes]: Note that $\mathbb{R}/\mathbb{Z}=\lbrace [r] : r\in \mathbb{R}\rbrace$, where $[r]=r+\mathbb{Z}$.
As Prahlad Vaidyanathan suggests above, consider the map $$f:\mathbb{R}/\mathbb{Z}\to S^1, \\ \;\;\;\;\;\;\;\;\;\;[r]\mapsto e^{2\pi i r}$$
This is a group homomorphism, since $$f([r]+[s])=f((r+s)+\mathbb{Z})=e^{2\pi i (r+s)}=e^{2\pi i r}e^{2\pi i s}=f([r])f([s]).$$
It only remains to prove that the map above is a bijection, or, in other words, that $f$ is both an injection and a surjection, which is not hard.