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 $$\left({T^S, \oplus}\right)$$.