Abstract:
The set $\mathrm{Hom}(G,H)$ of all homomorphisms from a group $G$ to a group $H$ is a group with respect to the operation of pointwise products iff the images of any two such homomorphisms commute element-wise; in this case, the group is commutative. For finite $G$ and $H$, we study algebraic properties of this group and of the union $\mathrm{Im}(G,H)$ of the images of all homomorphisms from $G$ to $H$. Let $\exp(G)$ be the minimal positive integer $n$ such that $x^n=1$ for all $x\in G$, let $G'$ be the commutator subgroup of $G$, $q=\exp(G/G')$, and let $\Omega_q(H)$ be the subgroup of $H$ generated by all elements of order $q$. We obtain the following results.
If $\mathrm{Hom}(G,H)$ is a group, then $\Omega_q(H)$ is commutative and the groups $\mathrm{Hom}(G,H)$ and $\mathrm{Hom}(G/G',\Omega_q(H))$ are isomorphic. Conversely, if $\Omega_q(H)$ is commutative and $\phi(G')=\{1\}$ for all $\phi\in\mathrm{Hom}(G,H)$, then $\mathrm{Hom}(G,H)$ is a group.
If $\mathrm{Im}(G,H)$ is a subgroup of $H$, then it is endomorphically admissible in $H$.
If $G$ is a finite $p$-group such that $\exp(G)=\exp(G/G')=q$ and $H$ is a regular $p$-group, then $\mathrm{Im}(G,H)=\Omega_q(H)$.