Indexed Summation does not Change under Permutation

Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $a$ and $b$ be integers.

Let $\closedint a b$ be the integer interval between $a$ and $b$.

Let $f: \closedint a b \to \mathbb A$ be a mapping.

Let $\sigma: \closedint a b \to \closedint a b$ be a permutation.

Then we have an equality of indexed summations:


 * $\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = a}^b \map f {\map \sigma i}$

Outline of Proof
Using Indexed Summation over Translated Interval, we reduce to the case $a = 1$.

Because Symmetric Group on n Letters is Generated by Standard Cycle and Adjacent Transposition, it suffices to do the cases where:
 * $\map \sigma i = i$ for all $i \in \closedint 1 b$ except for two consecutive integers
 * $\sigma$ is the standard cycle $\begin {bmatrix} 1 & \ldots & b \end {bmatrix}$

Proof
Let $a > b$.

Then by definition of indexed summation, both sides are $0$.

Let $a \le b$.

Reduction to $a=1$
By Indexed Summation over Translated Interval:
 * $\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^{b - a + 1} \map f {j - a + 1}$
 * $\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^{b - a + 1} \map f {\map \sigma {j - a + 1} }$

Let $n = b - a + 1$.

Because $a \le b$, we have $n \ge 1$.

By Translation of Integer Interval is Bijection, the mapping $T : \closedint 1 n \to \closedint a b$ defined by:
 * $\map T i = i + a - 1$

is a bijection.

We have:
 * $\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^n \map f {\map T j}$
 * $\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^n \map f {\map \sigma {\map T j} }$

By Inverse of Bijection is Bijection and Composite of Bijections is Bijection, the composition $T^{-1} \circ \sigma \circ T$ is a permutation of $\closedint 1 n$.

Suppose we have settled the case $a = 1$.

Then:
 * $\displaystyle \sum_{j \mathop = 1}^n \map f {\map T j} = \sum_{j \mathop = 1}^n \map {\paren {f \circ T} \circ \paren {T^{-1} \circ \sigma \circ T} } j$

By Composition of Mappings is Associative, this equals:
 * $\displaystyle \sum_{j \mathop = 1}^n \paren {f \circ \sigma \circ T} j$

Thus:
 * $\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = a}^b \map f {\map \sigma i}$

It remains to investigate the case $a = 1$.

The case $a = 1$
It remains to show that
 * $\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$

for any permutation $\sigma$ of $\closedint 1 n$.

Let $S_n$ be the symmetric group on $n$ letters, that is, the group of permutations of $\closedint 1 n$.

Let $U \subset S_n$ be the subset of all permutations $\sigma$ satisfying
 * $\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$

for all mappings $f: \closedint 1 n \to \mathbb A$.

Summation-preserving permutations form subgroup
We prove that $U$ is a subgroup of $S_n$, using Two-Step Subgroup Test.

We have to show that $U$ is non-empty.

By Identity Mapping is Permutation the identity mapping $I$ on $\closedint 1 n$ is a permutation.

By Identity Mapping is Right Identity, $f \circ I = f$.

Thus:
 * $\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map I i}$

Thus $I \in U$.

Let $\sigma, \tau \in U$.

We have to show that $\sigma \circ \tau$ is in $U$.

We have:

Thus $\sigma \circ \tau \in U$.

We show that $\sigma^{-1} \in U$.

We have:

Thus $\sigma^{-1} \in U$.

Thus $U$ is a subgroup of $S_n$.

Last adjacent transposition preserves summations
If $n = 1$, then by First Symmetric Group is Trivial Group and Subgroups of Trivial Group, $U = S_n$.

Let $n \ge 2$.

By Symmetric Group on n Letters is Generated by Standard Cycle and Adjacent Transposition, it suffices to show that $U$ contains an adjacent transposition and the standard cycle.

Let $\sigma \in S_n$ be the adjacent transposition $\begin {bmatrix} n - 1 & n \end {bmatrix}$.

We have:

Thus $\begin {bmatrix} n - 1 & n \end {bmatrix} \in U$.

Standard cycle preserves summations
Let $\sigma$ be the standard cycle $\begin {bmatrix} 1 & \ldots & n \end {bmatrix}$.

We have:

Thus $\begin {bmatrix} 1 & \ldots & n \end {bmatrix} \in U$.

Also see

 * Change of Variables in Indexed Summation

Generalizations

 * Indexed Iterated Operation does not Change under Permutation