how to prove multiplicative associative law for group semiring of set of functions from a finite group to semiring

Question

Let $S$ be a semiring and $(G,.)$ be a finite group. The set of functions $G \rightarrow S$ is a semiring with operations defined by $(f+g)a=f(a)+f(b)$ and $(fg)a=\displaystyle\sum_{bc=a}f(b)f(c)=\displaystyle\sum_{b \in G}f(b)f(b^{-1}a)$ for all $f,g:G \rightarrow S$ and $a \in G$. Anybody help me to prove multiplicative associative law for semiring.

Answer 1

Easiest to do it using the first (symmetric) form of the sum (for clarification, I use $\color{blue}*$ for multiplication in $S$, $\color{green}*$ for multiplication in $G$, and $\color{red}*$ for the multiplication of functions): $$\begin{align}(f\color{red}* (g\color{red}* h))(a)&=\sum_{b\color{green}* c=a}f(b)\color{blue}*(g\color{red}* h)(c)\\ &=\sum_{b\color{green}* c=a}f(b)\color{blue}*\sum_{d\color{green}* e=c}g(d)\color{blue}* h(e) \\ &=\sum_{b\color{green}* (d\color{green}* e)=a}f(b)\color{blue}*(g(d)\color{blue}* h(e))\end{align}$$ and similarly we find $$((f\color{red}* g)\color{red}* h)(a)=\ldots=\sum_{(b\color{green}* d)\color{green}* e=a}(f(b)\color{blue}*g(d))\color{blue}* h(e). $$ The last expressions uses only $\color{green}*$ and $\color{blue}*$, which we already know are associative, hence they are equal as desired.