# Show that $\mathbb{R}/\mathbb{Z}\cong U$ where $U=\{z\in \mathbb{C}\mid |z|=1\}$

Consider that $\mathbb{R}$ is a group under addition and $\mathbb{Z}$ is a subgroup of $\mathbb{R}$. Show that $\mathbb{R}/\mathbb{Z}\cong U$ where $U=\{z\in \mathbb{C}\mid |z|=1\}$

Consider the function $f:\mathbb{R}\rightarrow\mathbb{C}$ given by $f(x)=e^{2\pi xi}$. Since $\mathbb{R}$ is a additive group, then the additive identity $0\in \mathbb{R}$, so $f(0)=e^{2\pi 0i}=e^0=1\in \mathbb{C}$ which show $f(x)$ is not a trival function. Let $a,b\in \mathbb{R}$, then $f(a)=e^{2\pi ai}$ and $f(b)=e^{2\pi bi}$, so $f(a+b)=e^{2\pi (a+b)i}=e^{2\pi (ai+bi)}=e^{2\pi ai}e^{2\pi bi}=f(a)f(b)$. Hence, $f(x)$ is a homomorphism. Because $\mathbb{R}$ is an additive group,$\ker(f(x))=\{0\}$, by the first isomorphic theorem, $\mathbb{R}/\mathbb{Z}\cong U$.

Is that right? I am not sure the argument for kernel is right? can anyone show me or give me a hit to write a better proof? Thanks.

• $\operatorname{Ker}(f) = \Bbb Z$, not $\{0\}$ – jkabrg Jun 6 '15 at 22:33
• Your strategy is exactly right. – jkabrg Jun 6 '15 at 22:48

$Ker f = \lbrace x \in \mathbf{R}$ such that $e^{2i\pi x}=1 \rbrace =\mathbf{Z}$. Now you need to prove that the homomorphism is surjective on the unit disk, which you impicitely did!
• $f(x)$ is onto by construction because $\mathbb{R}$ is subset of $\mathbb{C}$? – Simple Jun 6 '15 at 23:03
• Nope. The unit disk is characterised by $\lbrace \theta \in \mathbf{R}, e^{i \theta} \rbrace$. To see that, you can use the polar form of a complex number. – mich95 Jun 6 '15 at 23:05
• Any $z \in \mathbb{C}$ can be written $z=re^{i \theta}$, where $r>0$ and $\theta \in [0,2 \pi[$.Now if $\vert z \vert=1$, then $r=1$. – mich95 Jun 6 '15 at 23:27