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Let $G$ be an infinite group and $x \in R[G]$ the group ring of $G$ over $R$. If $gx-x=0$ $\forall g \in G, g\neq 1$, then follows $x=0$.

I need a nice proof. I already have one dealing with the 'support of x'. But this just something I can imagine it to be true and nothing to write down.

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What's wrong with the proof using the support of x? It is very simple and intuitive. – studiosus Sep 18 '13 at 17:35
There is nothing wrong with the support argument. If $f$ is a $G$-invariant function then it is constant, and if $G$ is infinite while $f$ has finite support then we may easily argue $f\equiv0$ identically. As studiosus says, this is very simple and it's the idea that jumps out immediately. – anon Sep 18 '13 at 17:43
up vote 2 down vote accepted

Ok, supposing (as anon suggested above) that you are requiring that $\forall g \in G$ should holds $gx - x = 0$ then there's a direct proof.

We have that $x= \sum_{g' \in G} r_{g'} g'$, where the $r_{g'} \in R$ are all but a finite number equal to $0$.

By hypothesis for every $g \in G$ we have $$\sum_{g' \in G} r_{g'} gg' = gx = x = \sum_{g' \in G} r_{g'} g'$$ if $g g' = g''$ for some $g'' \in G$ this tells that $r_{g'} = r_{g''}$.

Because for every pair $g', g'' \in G$ there's a $g \in G$ such that $g g' = g''$ we can conclude that for every pair $g',g'' \in G$ we should have $r_{g'} = r_{g''}$ , but we have that all $r_g$ but a finite number a null (and the $r_g$ in the decomposition of $x$ are infinite since $G$ is infinite, so there must be a $r_g=0$), so they must all be null.

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Pretty sure it's $\forall g\in G$ not $\forall g\in R[G]$... – anon Sep 18 '13 at 17:40
@anon where is written that $g \in G$? I don't see in the question. :) Anyway I'm going to edit the answer soon. – Giorgio Mossa Sep 18 '13 at 17:52
It doesn't say which of $R[G]$ or $G$ it's in, so we need to infer from context clues which it must be. Since (a) it used the letter $g$ instead of something else, (b) you know the question is way too trivial if it's $\forall g\in R[G]$, and (c) you know the question called $G$ an infinite group and mentioned that a proof would use supports, you know the reasonable interpretation must be $\forall g\in G$. – anon Sep 18 '13 at 20:36

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