A finite sum in a commutative monoid indexed by an arbitrary finite set

Let $(M,+,0)$ be a commutative monoid (Semigroup and monoid), let $S$ be a finite set with $n:=\mathrm{\mid }S\mathrm{\mid }$ (The cardinality $\mid A\mid$ of a finite set), and let $a:S\to M$. Choose a bijection $\varphi :n\to S$ and define

${\sum }_{s\in S}{a}_s:=a_{\varphi (0)}+a_{\varphi (1)}+\cdots +a_{\varphi (n-1)},$

where the right side is the finite monoid product of The product $g_0{g}_1\cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n=0$) equal to the identity written additively. In particular, the sum over the empty set is $0$.

This value is independent of the enumeration. If $\psi :n\to S$ is another bijection, then ${\varphi }^{-1}\circ \psi$ is a permutation of $n$ (Injection, surjection, bijection), and generalised commutativity gives

${\sum }_{k<n}{a}_{\psi (k)}={\sum }_{k<n}{a}_{\varphi (k)}$

(Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either). Thus the displayed notation names one element of $M$ and does not select a preferred enumeration.

When $M$ is the additive monoid of $\mathbb{R}$ or of $\mathbb{N}$, this definition agrees with The sum ${\sum }_{i\in S}{a}_i$ over a finite index set, and its product form: both enumerate $S$, apply the same finite recursion with identity $0$, and are independent of the enumeration.

The coefficient object is an arbitrary commutative monoid. This is stronger than the published real- and natural-valued definition and is the form needed for sums in a commutative ring.