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Let $H$ be the subgroup of all elements of finite order in group $\left(\mathbb{C} \setminus \{0\}, \cdot \right)$. Prove that $H$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$, where $\mathbb{Q}$ and $\mathbb{Z}$ are groups with addition operation.

Do I need to work with mapping $\phi$ and its $\ker\phi$ and $\operatorname{Im} \phi$?

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    $\begingroup$ What mapping $\phi$ are you talking about? $\endgroup$ – Arthur Apr 16 at 10:24
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    $\begingroup$ @Arthur $\phi : H \rightarrow \mathbb{Q}/\mathbb{Z}$ $\endgroup$ – Nikita Gubanov Apr 16 at 10:27
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    $\begingroup$ I can think of several such maps. Which one are you talking about? How is it defined? $\endgroup$ – Arthur Apr 16 at 10:31
  • $\begingroup$ @Arthur From what I understand the map firstly has to be homomorphic, i.e $\phi(ab) = \phi(a) \cdot \phi(b)$ for any $a, b \in G$. Secondly, to be isomorphic it has to be bijective. And I guess we need to prove that, possibly by working with Ker and Im? $\endgroup$ – Nikita Gubanov Apr 16 at 10:38
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    $\begingroup$ Before you can prove that it's bijective, you have to know what it is. Do you have a formula, or a similar description of exactly what $\phi$ is? $\endgroup$ – Arthur Apr 16 at 10:41
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Obviously the elements of finite order are precisely those of the form form $z(q) = e^{2\pi i q}$ with $q \in \mathbb Q$. If $q = \frac{m}{n}$, then $z(q)^n = 1$, and if $z = re^{2\pi i t}$ has finite order, then $r^n e^{2\pi i nt} = 1$ for some $n > 0$ which implies $r = 1$ and $nt \in \mathbb Z$, i.e. $t \in \mathbb Q$.

Define $\phi : \mathbb Q \to \mathbb C^* = \mathbb C \setminus \{ 0 \}, \phi(q) = e^{2\pi i q}$. Obviously $\phi$ is a homomorphism such that $\mathrm{im}(\phi) = H$.

We have $\phi(q) = 1$ if and only if $q \in \mathbb Z$. Thus $\ker(\phi) = \mathbb Z$ and $\phi$ induces an isomomorphism $\phi' : \mathbb Q /\mathbb Z \to H$.

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