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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.

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

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  • $\begingroup$ $f(x)$ is onto by construction because $\mathbb{R}$ is subset of $\mathbb{C}$? $\endgroup$ – Simple Jun 6 '15 at 23:03
  • $\begingroup$ 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. $\endgroup$ – mich95 Jun 6 '15 at 23:05
  • $\begingroup$ 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 $r=1$. $\endgroup$ – mich95 Jun 6 '15 at 23:27
  • $\begingroup$ Is it convincing? $\endgroup$ – mich95 Jun 7 '15 at 1:06
  • $\begingroup$ yes, that is very helpful. $\endgroup$ – Simple Jun 7 '15 at 1:11

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