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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!

Martin Sleziak
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EgoKilla
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2 Answers2

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Hint:

Define

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

DonAntonio
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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 $\Bbb R / \Bbb Z$.

EgoKilla
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    If I understand your post correctly, you have found a group isomorphism between $S^1$ and $\mathbb R/\mathbb Z$. (The same was also shown in this answer.) Since you are interested in topological groups, you should also show that this map is a homeomorphism. This post might be interesting for you: http://math.stackexchange.com/questions/415735/first-isomorphism-theorem-for-topological-groups – Martin Sleziak Feb 24 '14 at 07:46