Let ##G## be a group and let ##R## be the set of reals.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Convolution on groups

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