# Finite Submodule of Function Space

## Theorem

Let $\struct {G, +}$ be a group whose identity is $e$.

Let $R$ be a ring.

Let $\struct {G, +, \circ}_R$ be an $R$-module.

Let $S$ be a set.

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

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

Then:

- $\struct {G^{\paren S}, +', \circ}_R$ is a submodule of $\struct {G^S, +, \circ}_R$

where $+'$ is the operation induced on $G^{\paren S}$ by $+$.

## Proof

Let $\struct {G, +, \circ}_R$ be an $R$-module and $S$ be a set.

We need to show that $\struct {G^{\paren S}, +'}$ is a group.

Let $f, g \in G^{\paren S}$.

Let:

- $F = \set {x \in S: \map f x \ne e}$
- $G = \set {x \in S: \map g x \ne e}$

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:

- $\map f x + \map g x \ne e$

then necessarily $\map f x \ne e$ or $\map g x \ne e$.

That is:

- $\map {\paren {f +' g} } x = \map f x + \map g x \ne e \implies x \in F \cup G$

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

Hence $f +' g \in G^{\paren S}$ and $\struct {G^{\paren S}, +'}$ is closed.

Now let $f \in G^{\paren S}$.

Let $f^*$ be the pointwise inverse of $f$.

Thus:

- $\map {f^*} x = -\paren {\map f x}$

Again, let $F = \set {x \in S: \map f x \ne e}$.

It follows directly that:

- $x \in S \setminus F \implies \map {f^*} x = e$

Hence:

- $\map {f^*} x \ne e \implies x \in F$

and hence:

- $f^* \in G^{\paren S}$

So by the Two-Step Subgroup Test, it follows that $\struct {G^{\paren S}, +'}$ is a subgroup of $\struct {G^S, +}$.

Hence $\struct {G^{\paren S}, +', \circ}_R$ is an $R$-module.

The result follows.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.3$