What topological group is $\mathbb R/\mathbb Z$?

Question

The integers $\mathbb Z$ are a normal subgroup of $(\mathbb R, +)$. The quotient $\mathbb R/\mathbb Z$ is a familiar topological group; what is it?

I've found elsewhere on the internet that it is the same as the topological group $(S^1, *)$ but have no idea how to show this. Any help would be appreciated.

Thanks!

Answer 1

After some work here's my answer.

Begin with: $$\phi:\Bbb R\to S^1\le\Bbb C^*\;,\;\;\phi(r):=e^{2\pi i r}$$

Then we want to show that $\phi$ is a group homomorphism between $\Bbb R$ and $S^1$.

Note: By Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$.

First we prove that $\phi$ is surjective. This is easy to see since from our note $\sin$ and $\cos$ must span all possible values of a and b to give norm 1.

Again from our note it is clear that $\phi$ maps all integers to 1, the identity element of $S^1$. This is precisely $\ker(\phi)$.

Thus, by the first isomorphism theorem we have $(S^1, *)$ isomorphic to $\Bbb R / \Bbb Z$.

Answer 2

Hint:

Define

$$\phi:\Bbb R\to S^1\le\Bbb C^*\;,\;\;\phi(r):=e^{2\pi i r}$$