Finite Submodule of Function Space

Theorem
Let $$G$$ 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 $$\left({G^S: \circ}\right)_R$$.

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

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

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

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 $$x \in S - F \Longrightarrow f^* \left({x}\right) = e$$.

Hence $$f^* \left({x}\right) \ne e \Longrightarrow x \in F$$ 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.