Exercise 2.3.19 (Universal property of $\mathbb{Z}$)

Proof. Let $H$ a group and $h\in H$.

• Unicity. If $f:\mathbb{Z}\to H$ is a homomorphism such that $f(1)=h$, then for all $k\in \mathbb{Z}$, $f(k)=h^k$ (proof similar to that of the Exercise 18). If $g$ is another such homomorphism, then for all $k\in \mathbb{Z}$,

  $$g(k)=h^k=f(k),$$

  so $f=g$.

• Existence. Consider the map

  $$f\{\begin{array}{ccc}\hfill \mathbb{Z}\hfill & \hfill \to \hfill & H\hfill \\ \hfill k\hfill & \hfill \mapsto \hfill & h^k\hfill \end{array}$$

  Then, for all $k,l\in \mathbb{Z}$,

  $$f(k+l)=h^{k+l}=h^k{h}^l=f(k)f(l),$$

  so $f$ is a homomorphism. Moreover $f(1)=h^1=h$.

In conclusion, if $H$ is any group and $h$ is an element of $H$, then there is a unique homomorphism from $\mathbb{Z}$ to $H$ such that $1\mapsto H$. □