Can the Set R(G) be Proven as a Ring in Convolution on Groups?

[Bipolarity]

Let $G$ be a group and let $R$ be the set of reals.
Consider the set $R(G) = \{ f : G \rightarrow R \, | f(a) \neq 0$ for finitely many $a \in G \}$.
For $f, g \in R(G)$, define $(f+g)(a) = f(a) + g(a)$ and $(f * g)(a) = \sum_{b \in G} f(b)g(b^{-1}a)$.

Prove that $R(G)$ is a ring.
I'm not even sure if this is true, so I'm posting it here. I proved everything except closure under product, which is giving me some trouble. To show closure under product, I'd need to show

$\sum_{b \in G} f(b)g(b^{-1}a) \neq 0$ is true for finitely many $a \in G$. But is this even true??

Thanks!

BiP

 

[jbunniii]

First, let's check that the sum defining the convolution is well defined. Since $G$ is presumably infinite (otherwise the problem is not very interesting), and $f,g$ are not constrained to be nonnegative, the sum $\sum_{b \in G}$ is not well defined unless only finitely many of the terms are nonzero.

We're given that $f(b)$ is nonzero for only finitely many $b \in G$. Also, for a fixed value of $a$, we are also given that $g(b^{-1}a)$ is nonzero for only finitely many $b \in G$.

Therefore, for fixed $a$, the product $f(b)g(b^{-1}a)$ is nonzero for only finitely many $b \in G$. So the sum $\sum_{b \in G}f(b)g(b^{-1}a)$ makes sense for every $a \in G$, and the convolution is well defined.

Let's define $N(f) = \{x \in G \mid f(x) \neq 0\}$ and $N(g) = \{x \in G \mid g(x) \neq 0\}$. Both sets are finite since $f,g \in R(G)$.

Now, for a given $a \in G$, if $(f * g)(a) = \sum_{b \in G}f(b)g(b^{-1}a)$ is nonzero, then there must be at least one $b \in G$ such that both $f(b)$ and $g(b^{-1}a)$ are nonzero. So $b \in N(f)$ and $b^{-1}a \in N(g)$. The condition $b^{-1}a \in N(g)$ is equivalent to $a \in bN(g)$.

The two conditions $b\in N(f)$ and $a \in bN(g)$ together imply that $a \in N(f)N(g)$, and the latter is a finite set.