# Prove that subgroup of all elements of finite order in group $(\mathbb{C} \setminus \{0\}, \cdot)$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$

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 $$\operatorname{Im} \phi$$?

• What mapping $\phi$ are you talking about? – Arthur Apr 16 at 10:24
• @Arthur $\phi : H \rightarrow \mathbb{Q}/\mathbb{Z}$ – Nikita Gubanov Apr 16 at 10:27
• I can think of several such maps. Which one are you talking about? How is it defined? – Arthur Apr 16 at 10:31
• @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? – Nikita Gubanov Apr 16 at 10:38
• 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? – Arthur Apr 16 at 10:41

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 $$\phi' : \mathbb Q /\mathbb Z \to H$$.