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 $\struct {T, \oplus}$.

Group Axiom $\text G 0$: Closure

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

Hence a fortiori $\struct {T, \oplus}$ 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.

Group Axiom $\text G 3$: Existence of Inverse Element

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 $\struct {T^S, \oplus}$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.7$: Corollary