Set of Homomorphisms to Abelian Group is Subgroup of All Mappings

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure.

Let $\left({T, \oplus}\right)$ be an abelian group.

Let $\left({T^S, \oplus}\right)$ be the algebraic structure on $T^S$ induced by $\oplus$.

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

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

G0: Closure
From Homomorphism on Induced Structure, we have that $\forall f, g \in H: f \oplus g$ is a homomorphism from $\left({S, \circ}\right)$ into $\left({T, \oplus}\right)$.

Hence $\left({H, \oplus}\right)$ is closed.

G3: Inverses
From Inverse Mapping in Induced Structure, if $g$ is a homomorphism then its Induced Structure Inverse $g^*$ is one also.

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

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