Help with $\operatorname{Hom}(\langle x \rangle, \mu) \cong \mu_{n}$

Question

Let $G$ be a group and $x \in G$ with $|x|=k$.

Let $\mu_n:=\{z \in \mathbb{C} \mid z^n=1\}$ and $\mu=\bigcup_{n=1}^\infty \mu_n$.

I want to show that $\operatorname{Hom}(\langle x \rangle, \mu) \cong \mu_k$. How do I identify elements of the LHS with elements of the RHS?

Answer 1

Define the evaluation in $x$ $$ev_x :\operatorname{Hom}(\langle x \rangle, \mu) \longrightarrow \mu_n \ \ \ , \ \ \ f \mapsto f(x)$$

Note that this is well defined, since $(f(x))^n = f(x^n) = f(1) = 1$.

The kernel is trivial since $\forall f \in \ker (ev_x), \forall k $, you have $f(x^k) = (f(x))^k = (ev_x(f))^k = 1^k = 1$.

Finally $ev_x$ is surjective since $\forall z \in \mu_n$ you can define $f : \langle x \rangle \longrightarrow \mu$ $$f(x^k) = z^k$$ (it is well defined!) and you have that $f \in \operatorname{Hom}(\langle x \rangle, \mu) $ and $ev_x(f) = z$.

Answer 2

It's sufficient to take $G = \langle{x\rangle} = \mathbb{Z}_n$. Since $G$ is cyclic of order $n$ with generator $g$, we have a bijection between $\text{Hom}(G, X)$ and the set of elements of $X$ of order dividing $n$ defined by $f \to f(x)$. It's easy to verify that this map is also a homomorphism. Since $\mu_n$ is exactly the set of elements of $\mu$ of order $n$, we have the required isomorphism $\text{Hom}(G, \mu) = \mu_n$.