Finite Submodule of Function Space

Theorem
Let $\left({G, +}\right)$ be a group whose identity is $e$.

Let $R$ be a ring.

Let $\left({G, +, \circ}\right)_R$ be an $R$-module.

Let $S$ be a set.

Let $G^S$ the set of all mappings $f: S \to G$.

Let $G^{\left({S}\right)}$ be the set of all mappings $f: S \to G$ such that $f \left({x}\right) = e$ for all but finitely many elements $x$ of $S$.

Then:
 * $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is a submodule of $\left({G^S, +', \circ}\right)_R$

where $+'$ is the operation induced on $G^S$ by $+$.

Proof
Let $\left({G, +, \circ}\right)_R$ be an $R$-module and $S$ be a set.

We need to show that $\left({G^{\left({S}\right)}, +'}\right)$ is a group.

Let $f, g \in G^{\left({S}\right)}$.

Let:
 * $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$
 * $G = \left\{{x \in S: g \left({x}\right) \ne e}\right\}$

From the definition of $f$ and $g$, both $F$ and $G$ are finite.

Then:
 * $\forall x \in S \setminus F: f \left({x}\right) = e$ and so $f \left({x}\right) + g \left({x}\right) = e$
 * $\forall x \in S \setminus G: g \left({x}\right) = e$ and so $f \left({x}\right) + g \left({x}\right) = e$

So:
 * $\left({f +' g}\right) \left({x}\right) = f \left({x}\right) + g \left({x}\right) \ne e \implies x \in F \cap G$

But as $F$ and $G$ are both finite, it follows that $F \cap G$ is also finite.

Hence $f +' g \in G^{\left({S}\right)}$ and $\left({G^{\left({S}\right)}, +'}\right)$ is closed.

Now let $f \in G^{\left({S}\right)}$.

Let $f^*$ be the Induced Structure Inverse of $f$.

Thus $f^* \left({x}\right) = - \left({f \left({x}\right)}\right)$.

Again, let $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$.

It follows directly that $x \in S - F \implies f^* \left({x}\right) = e$.

Hence $f^* \left({x}\right) \ne e \implies x \in F$ and hence $f^* \in G^{\left({S}\right)}$.

So by the Two-Step Subgroup Test, it follows that $\left({G^{\left({S}\right)}, +'}\right)$ and hence $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is an $R$-module.

The result follows.