Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.19
Show that if $H$ is a group and $h \in H$, there exists a unique homomorphism $\varphi : \mathbb{Z} \rightarrow H$ such that $\varphi(1) = h$.
Solution:
Existence: Define $\varphi(n) = h^n$. We need not worry about well-definedness for $\varphi$. For all $m,n \in \mathbb{Z}$, we have $$\varphi(m+n) = h^{m+n} = h^mh^n = \varphi(m)\varphi(n),$$ so that $\varphi$ is a homomorphism.
Uniqueness: Suppose we have another homomorphism $\psi$ such that $\psi(1) = h$. Then $$\psi(n) = \psi(n \cdot 1) = \psi(1)^n = \varphi(1)^n = \varphi(n),$$ so that $\psi = \varphi$ and so $\varphi$ is unique. (Keep in mind that we write $Z$ additively and $H$ multiplicatively.)
