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This actually causes a lot of confusion. For a finite group $G$ and a commutative unit ring $R$ I’m trying to prove that $h\in R[G]$ defined as $h=\sum\limits_{g\in G}g$ is in the center of $R[G]$. How do you write $h$ as an element of $Maps(G,R)$? How do you interpret the coefficients $c_g$ in $R[G]=\{\sum\limits_{g\in G} c_g\cdot g \ :\ c_g\in R\}$ relative to an $h\in R[G]$?

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    $\begingroup$ When rewriting a function of finite support as a formal linear combination, the coefficient of an element $g$ in the linear combination is the value in $R$ on that $g$ of the corresponding function. So $h$ has coefficient $1$ for each $g\in G$, meaning $h$ is the function $g\mapsto 1$ for all $g\in G$. $\endgroup$ – Ben West Feb 21 '18 at 16:51
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The notation $f_g$ is just an alternative functional notation for $f(g)$. That's what the comment is referring to.

The element $h=\sum_{g\in G} g=\sum _{g\in G}1\cdot g$ can be interpreted as a map $G\to R$ by saying that $g\mapsto 1$ for all $g\in G$.

The coefficients are simply a listing of where the map sends each $g\in G$. So if you know the coefficients, you know the map, and if you know the map you know the coefficients.

Is it absolutely necessary to prove it's in the center via the map interpretation? Isn't it much easier to notice that multiplying $gh=h=hg$ for all $g\in G$? Then since $h$ commutes with everything in $R$ and $G$, it commutes with everything in $R[G]$.

(Personally I've never got any use out of the functional interpretation, but then again I don't have a lot of representation theory training.)

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  • $\begingroup$ So $h$ is a map that sends everything to $1$ but it is also the sum of all the elements of $G$ multiplied by $1$ on the left... so which is it? $\endgroup$ – John Cataldo Feb 21 '18 at 17:10
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    $\begingroup$ @StanislasHildebrandt it is the exact same, from two different formal points of view. $\endgroup$ – Arnaud Mortier Feb 21 '18 at 17:15
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    $\begingroup$ @StanislasHildebrandt just like a complex number $a+ib$ can be seen as a function $\{1, i\}\to \Bbb R$, mapping $1$ to $a$ and $i$ to $b$. $\endgroup$ – Arnaud Mortier Feb 21 '18 at 17:17
  • $\begingroup$ @StanislasHildebrandt The beauty of interpretations is that it doesn't have to be one or the other. It can be both, and you learn different things about it that way. $\endgroup$ – rschwieb Feb 21 '18 at 17:24
  • $\begingroup$ @ArnaudMortier, you are recommending (as I do also) the fourfold vision in mathematics. The more different ways we know for viewing a mathematical phenomenon, the better we can understand it. $\endgroup$ – Lubin Feb 21 '18 at 18:28
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How do you interpret the coefficients $c_g$ in $R[G]=\{\sum\limits_{g\in G} c_g\cdot g \ :\ c_g\in R\}$ relative to an $h\in R[G]$?

The coefficients are part of the definition of a group ring. A group ring $R[G]$ is the set of linear combinations of elements of the group, with coefficients in the ring, endowed with the natural operations (notably the product in $R[G]$ combines the product in $R$ and that in $G$).

$$h=\sum\limits_{g\in G}g$$ How do you write $h$ as an element of $Maps(G,R)$?

Each element of $R[G]$ is characterised by the coefficient in front of each element of $G$. Therefore $\sum_\limits{g\in G}c_g\cdot g$ can be thought of as the map $g\mapsto c_g$.

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