Show that ${\rm Hom}(G,\mathbb{C}^{\star}) \cong G \cong {\rm Hom}(G,\mathbb{Q}/\mathbb{Z})$ How can I show that ${\rm Hom}(G,\mathbb{C}^{\star}) \cong G \cong {\rm Hom}(G,\mathbb{Q}/\mathbb{Z})$, where $G$ is finite abelian group, $\mathbb{C}$ is the set of complex number, $\mathbb{Q}$ is the set of rational, $\mathbb{Z}$ is the set of integers and ${\rm Hom}(A,B)$ is the set of all homomorphisms from a group $A$ to a group $B$.
 A: $\newcommand{Hom}{\text{Hom}}$
Since $G$ is a finite direct sum of finite cyclic groups, and $\Hom$ commutes with finite direct sums (in either variable), it suffices to show the result for $G = \mathbb{Z}/n\mathbb{Z}$.
The key fact is that for any $n \in \mathbb{N}$, both $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{C}^*$ contain a unique cyclic subgroup of order $n$ (for $\mathbb{Q}/\mathbb{Z}$, a generator is $\frac{1}{n} + \mathbb{Z}$, and for $\mathbb{C}^*$, a generator is a primitive $n^\text{th}$ root of unity). Thus e.g. if $H \subseteq \mathbb{Q}/\mathbb{Z}$ is the cyclic subgroup of order $n$, then any homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Q}/\mathbb{Z}$ must land in $H$, i.e. $\Hom(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) = \Hom(\mathbb{Z}/n\mathbb{Z}, H)$. But $H \cong \mathbb{Z}/n\mathbb{Z}$, so $\Hom(\mathbb{Z}/n\mathbb{Z}, H) \cong \Hom(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/n\mathbb{Z}$.
A: First you have to define a group action, say $\ast$, on the set $Hom(G, \mathbb{C}^\ast)$ as follows:
$$
(\rho \ast \phi) (g):= \rho(g) \phi(g), \quad \forall \rho, \phi \in Hom(G, \mathbb{C}^\ast), g\in G,
$$
where the multiplication in the right is the multiplication in $\mathbb{C}$. Then since $G$ is a finite group you see that every element of $Hom(G, \mathbb{C}^\ast)$ is of finite order, and so lies in the unit circle $\mathbb{T}$. On the other hand the subgroup of all finite order elements of $\mathbb{T}$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$.   
To prove the first isomorphism, note that since $G$ is a finite abelian group you can write it in the form $G\simeq \mathbb{Z}/n_1 \mathbb{Z}\oplus \cdots, \oplus \mathbb{Z}/n_k \mathbb{Z}$ for some positive integers $n_1,\cdots, n_k$. Then it is easy to define the first isomorphism using the generators of the right hand side. For example let $\phi$ be the homomorphism sending the generator of $\mathbb{Z}/n_1 \mathbb{Z}$ to $e^{2\pi /n_1}$. 
A: Ideas for the first:
$${\rm Hom}(C_n,{\Bbb C}^*) \cong  {\rm Hom}(C_n,C_n) = C_n,$$
$${\rm Hom}(G\times H,A) = {\rm Hom}(G,A)\times {\rm Hom}(H,A).$$
Where $C_n=$cyclic group, $A=$ abelian group.
A: Embed $\mathbb Q/\mathbb Z$ into $\mathbb C^*$ by the composite map
$$ \mathbb Q/\mathbb Z \hookrightarrow \mathbb R/\mathbb Z \xrightarrow{\exp(\tau i -)} \mathbb C^* $$
It is easy to see that an element of finite order in $\mathbb C^*$ must lie in $\mathbb Q/\mathbb Z$, so the first statement is equivalent to the second. By the fundamental theorem of finite abelian groups and the fact that $\operatorname{Hom}$ converts coproducts in the first variable to products, it suffices to prove the statement for finite cyclic groups, which is easy.
