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Sep 1, 2015 at 12:36 comment added Geoff Robinson Yes, thanks, that's true for $I(G)$ itself- not sure if that makes it clear that the element lies in the intersection of the powers of $I(G)$. I had noticed that that direction works (with my argument) for infinite groups too since $I(G)$ is still an ideal ( if we think of the group algebra as consisting of finitely supported elements)
Sep 1, 2015 at 11:51 comment added Ilya Bogdanov One may simplify the opposite direction argument by observing that $\mathbb F[G]/I(G)\cong \mathbb F$ via the epimorphism $\sum_{g\in G}\alpha_gg\mapsto \sum_{g\in G}\alpha_g$. This applies to all groups, not only finite ones.
Sep 1, 2015 at 11:37 history edited Geoff Robinson CC BY-SA 3.0
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Sep 1, 2015 at 10:19 history answered Geoff Robinson CC BY-SA 3.0