# 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 $G^{\left({S}\right)}$ 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.

Since $e + e = e$ by definition of identity element, it follows that if:

$f \left({x}\right) + g \left({x}\right) \ne e$

then necessarily $f \left({x}\right) \ne e$ or $g \left({x}\right) \ne e$.

That is:

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

But as $F$ and $G$ are both finite, it follows that $F \cup G$ 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 $x \in S \setminus F \implies f^* \left({x}\right) = e$.

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 $\left({G^{\left({S}\right)}, +'}\right)$ is a subgroup of $\left({G^S, +}\right)$.

Hence $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is an $R$-module.

The result follows.

$\blacksquare$