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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Convolution on groups

**Physics Forums | Science Articles, Homework Help, Discussion**