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.
Note. This does not really answer the question: I was assuming that $\mathbb C^*$ meant the multiplicative group of complex numbers, as it usually does, and not the multiplicative group of compex numbers of modulus $1$, as it did here in the first version of the question.
The abelian group $\mathbb R$ is isomorphic to $\mathbb R\oplus\mathbb R$, and one can pick the isomorphism so that the subgroup $\mathbb Z$ is mapped to the subgroup $0\oplus\mathbb Z\subset\mathbb R\oplus\mathbb R$. It follows that $\mathbb R/\mathbb Z\cong\mathbb R\oplus\mathbb R/0\oplus\mathbb Z\cong\mathbb R\oplus(\mathbb R/\mathbb Z)$.
Now the map $(x,y)\in \mathbb R\oplus(\mathbb R/\mathbb Z)\mapsto e^{x+2\pi i y}\in\mathbb C^*$ is an isomorphism.
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.
