Set of Homomorphisms to Abelian Group is Subgroup of All Mappings

Theorem
Let $\struct {S, \circ}$ be an algebraic structure.

Let $\struct {T, \oplus}$ be an abelian group.

Let $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $\oplus$.

Then the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$.

Proof
Let $H$ be the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$.

We have that $\struct {T, \oplus}$ be an abelian group.

Hence $\struct {T, \oplus}$ is a commutative semigroup.

From Homomorphism on Induced Structure to Commutative Semigroup:
 * $\forall f, g \in H: f \oplus g$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$.

Hence $\struct {H, \oplus}$ is closed.

From Inverse Mapping in Induced Structure of Homomorphism to Abelian Group, if $g$ is a homomorphism then its pointwise inverse $g^*$ is one also.

Thus $g \in H \implies g^* \in H$.

So by the Two-Step Subgroup Test, $\struct {H, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$.