Isomorphism between $ \textbf{Grp}(\mathbb{Z}, G)$ and $G$

Question

I'm learning some category after taking a break from algebra.

Could someone explain to me why, for any Group $G$, the set of morphisms between $\mathbb{Z}$ and $G$ is isomorphic to the group itself, that is,

$$ \textbf{Grp}(\mathbb{Z}, G) \cong G$$

Answer 1

The group $\mathbb{Z} = \langle 1 \mid \;\;\; \rangle$ is the free group with one generator. That means we have a one-to-one correspondence between homomorphisms $\mathbb{Z} \rightarrow G$ and maps $\lbrace 1 \rbrace \rightarrow G$, the latter representing the choices of some element $g \in G$. In other words we get the map $\varphi \colon \textbf{Grp}(\mathbb{Z}, G) \rightarrow G$, $f \mapsto f(1)$ and on the other hand the map $\psi \colon G \rightarrow \textbf{Grp}(\mathbb{Z}, G)$, $g \mapsto (1 \mapsto g)$, which sends some element $g \in G$ to the unique homomorphism defined by the map $1 \mapsto g$ and explicitly given by $a \mapsto g^a$. By construction, we have $\varphi = \psi^{-1}$, which means that $\varphi$ (and of course also $\psi$) are bijections.

Recall, that the group structure on $\textbf{Grp}(\mathbb{Z}, G)$ is defined as $(f \cdot g)(a) = f(a) \cdot g(a)$, where the latter multiplication is the multiplication of $G$. That way we pass on the group structure of $G$ to $\textbf{Grp}(\mathbb{Z}, G)$. Now you can check that $\varphi$ actually not only is a bijection, but also a homomorphism of groups and thus an isomorphism.