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Intuitively, I see how is related to $\{e^{i\theta} : 0 \le \theta \le 2\pi \}$. I tired to use the first Isomorphism theory where it states that the image of φ is isomorphic to the quotient group G / ker(φ). Now, I am so confused now since the kernel doesn't work out. I am having a hard time to put everything together... How do I prove $\mathbb{R}/\mathbb{Z}$ is isomorphic to $e^{i\theta}$. Or if my intuition is wrong and its isomorphic to something else? I'd appreciate your help!

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Perhaps try the map $\theta \mapsto e^{2\pi i \theta}$ instead of $\theta \mapsto e^{i\theta}$. – Tom Nov 8 '13 at 2:10
up vote 4 down vote accepted

Let $f\colon\mathbb{R}\rightarrow S^1=\{e^{2i\pi\theta}\}$ be given by $f(t)=e^{2i\pi t}$ (show this is a homomorphism). The kernel of $f$ is $\{t\in\mathbb{R}\mid f(t)=1\}=\{t\in\mathbb{Z}\}$ and so $\ker f= \mathbb{Z}$. We also know that $f$ is clearly surjective and so by the first isomorphism theorem, we have $$\mathbb{R}/\mathbb{Z}=\mathbb{R}/{\ker f}\cong\mbox{Im }f=S^1.$$

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Since a real number $x$ ca be uniquely writen $x=k+m$ where $(k,m) \in \mathbb Z \times [0,1[$, we can define the map $m=\mu(x)$ from $\mathbb R$ to $[0,1[$. Under the law: $a \bot b = \mu(a+b)$ we can see that $H=([0,1[,\bot)$ is a group. Since $\mu$ is a morphism and $\text{Im} (\mu)=[0,1[$ we have ${\mathbb R}/{\mathbb Z} \sim H$

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