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Let $G$ be a group, $R$ be a ring. One can then define the groupring $RG = \{f: G \to R \mid \sup(g) \ \mathrm{is \ finite}\}$, with pointwise addition and with the following multiplication:

If $\alpha, \beta \in RG$, then the product is defined as the function defined by $$\alpha \beta(z) = \sum_{x,y \in G, xy = z} \alpha(x)\beta (y) = \sum_{x \in G} \alpha(x)\beta(x^{-1}z)$$

Why does the function $\alpha\beta$ have finite support? I can see that for any $z \in G$, the sum above is finite, so the sum is well defined, but can't seem to find an easy reason why $\sup(\alpha\beta)$ is finite.

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    $\begingroup$ You only get a nonzero result if $z=xy$ where $x$ and $y$ are in the supports of $\alpha$ and $\beta$. $\endgroup$ – Lord Shark the Unknown Jan 18 '18 at 11:08
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$supp(\alpha\beta)\subseteq supp(\alpha)supp(\beta)=\{xy\mid x\in supp(\alpha), y\in supp(\beta)\}$.

Therefore $|supp(\alpha\beta)|\leq |supp(\alpha)||supp(\beta)|<\infty$

Suppose $z\in supp(\alpha\beta)$. Since $\alpha\beta(z)$ is nonzero, then at so is $\alpha(x)\beta(x^{-1}z)$ for at least one $x$. For this to be true, both $\alpha(x)$ and $\beta(x^{-1}z)$ have to be nonzero, so that $x\in supp(\alpha)$ and $x^{-1}z\in supp(\beta)$. So finally, $z=xx^{-1}z\in supp(\alpha)supp(\beta)$.

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  • $\begingroup$ I didn't use $\sup$ because that's the standard notation for supremum in my mind. $\endgroup$ – rschwieb Jan 18 '18 at 17:26
  • $\begingroup$ Yeah, I did want to write supp first too but it doesn't format nicely with the backslash. $\endgroup$ – user370967 Jan 18 '18 at 18:39
  • $\begingroup$ Also, can you work out why the first inclusion you wrote is true? $\endgroup$ – user370967 Jan 18 '18 at 18:42
  • $\begingroup$ @Math_QED The computation is trivial: I added it. $\endgroup$ – rschwieb Jan 18 '18 at 19:06
  • $\begingroup$ "spt" is not built into (La)TeX, but is easily distinguished from "sup"... $\endgroup$ – paul garrett Jan 18 '18 at 19:39

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