as

**Swlabr **mentioned, that's just the universal property of a free group on one generator. but it doesn't mean that the proof is hard!

:

consider the maps $\displaystyle H \overset{\psi} \longrightarrow \mathbb{Z} \overset{\phi} \longrightarrow H,$ where $\displaystyle \psi$ is the homomorphism sending $\displaystyle h$ to $\displaystyle 1$ (your map) and $\displaystyle \phi$ is defined by $\displaystyle \phi(n)=h^n,$ for all $\displaystyle n \in \mathbb{Z}.$ note

that $\displaystyle \phi$ is

__one-to-one__. (why?) now consider the maps $\displaystyle \phi \psi : H \longrightarrow H$ and $\displaystyle \text{id}: H \longrightarrow H,$ the identity map. we have $\displaystyle \phi \psi(h)=\phi(1)=h=\text{id}(h).$

thus by the "uniqueness" property: $\displaystyle \phi \psi = \text{id}.$ therefore for any $\displaystyle x \in H$ we have $\displaystyle \phi(\psi(x))=x,$ i.e. $\displaystyle \phi$ is

__onto__ and the proof is complete.