Change of Variables in Summation over Finite Set

Theorem
Let $G$ be an abelian group.

Let $S$ and $T$ be finite sets.

Let $f : S \to G$ be a mapping.

Let $g: T\to S$ be a bijection.

Then we have an equality of summations over finite sets:


 * $\displaystyle \sum_{s \in S}f(s) = \sum_{t \in T} f(g(t))$

Outline of Proof
This follows from the definition of summation over finite set and the fact that Summation over Finite Set is Well-Defined.

Proof
Let $S$ be empty.

By Bijection with Empty Set implies Empty Set, $T$ is empty.

By definition of summation over empty set, both sides equal the identity element of $G$.

Let $S$ be nonempty.

Let $n$ be the cardinality of $S$.

Let $\N_{<n}$ be an initial segement of the natural numbers.

Let $h : \N_{<n} \to T$ be a bijection.

By definition of summation over finite set, $\displaystyle \sum_{t \mathop\in T} f(g(t)) = \displaystyle \sum_{i\mathop= 0}^{n-1} f(g(h(i)))$.

By Composite of Bijections is Bijection, the composition $g\circ h : \N_{<n} \to S$ is a bijection.

By definition of summation over finite set, $\displaystyle \sum_{s \mathop\in S} f(s) = \displaystyle \sum_{i\mathop= 0}^{n-1} f(g(h(i)))$.

Also see

 * Finite Summation is Invariant under Permutation