Sign of Permutation is Plus or Minus Unity

Theorem
Let $$S_n$$ denote the symmetric group on $n$ letters.

Let $$\left \langle {x_k} \right \rangle_{k \in \mathbb{N}^*_n}$$ be a sequence in $\mathbb{R}$.

Let $$\pi \in S_n$$.

Let $$\Delta_n \left({x_1, x_2, \ldots, x_n}\right)$$ be the product of differences of $$\left({x_1, x_2, \ldots, x_n}\right)$$.

Let $$\pi \cdot \Delta_n \left({x_1, x_2, \ldots, x_n}\right)$$ be defined as: $$ \pi \cdot \Delta_n \left({x_1, x_2, \ldots, x_n}\right) = \Delta_n \left({\pi\left({x_1}\right), \pi\left({x_2}\right), \ldots, \pi\left({x_n}\right)}\right)$$.

Then either:
 * 1) $$\pi \cdot \Delta_n = \Delta_n$$, or:
 * 2) $$\pi \cdot \Delta_n = -\Delta_n$$.

The sign of a permutation $$\pi \in S_n$$ is defined as:

$$\sgn \left({\pi}\right) = \frac {\Delta n} {\pi \cdot \Delta n}$$

That is:

$$\sgn \left({\pi}\right) = \begin{cases} 1 & :\pi \cdot \Delta n = \Delta n \\ -1 & : \pi \cdot \Delta n = -\Delta n \end{cases}$$

$$\sgn \left({\pi}\right)$$ is voiced "signum pi".

Thus $$\pi \cdot \Delta n = \sgn \left({\pi}\right) \Delta_n$$.

Proof

 * If $$\exists i, j \in \mathbb{N}^*_n$$ such that $$x_i = x_j$$, then $$\Delta_n \left({x_1, x_2, \ldots, x_n}\right) = 0$$ and the result follows trivially.

So, suppose all the elements $$x_k$$ are different.

Let us use $$\Delta_n$$ to denote $$\Delta_n \left({x_1, x_2, \ldots, x_n}\right)$$.


 * Let $$1 \le a < b \le n$$. Then $$x_a - x_b$$ is a divisor of $$\Delta_n$$.

Then $$x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}$$ is a factor of $$\pi \cdot \Delta_n$$.

There are two possibilities for the ordering of $$\pi \left({a}\right)$$ and $$\pi \left({b}\right)$$:

Either $$\pi \left({a}\right) < \pi \left({b}\right)$$ or $$\pi \left({a}\right) > \pi \left({b}\right)$$.

If the former, then $$x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}$$ is a factor of $$\Delta_n$$.

If the latter, then $$- \left({x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}}\right)$$ is a factor of $$\Delta_n$$.

The same applies to all factors of $$\Delta_n$$.

Thus:

$$\pi \cdot \Delta_n = \pi \cdot \prod_{1 \le i < j \le n} \left({x_i - x_j}\right) = \pm \prod_{1 \le i < j \le n} \left({x_i - x_j}\right) = \pm \Delta_n$$