The Sign of a Permutation is Multiplicative

Let $n$ be a natural number, let $[n]$ be the initial segment determined by $n$, and let $S_{n}$ be the set of permutations of $[n]$, with the identity $\mathrm{id}$, the composition $\sigma\circ\tau$ and the inverse $\sigma^{-1}$ of Permutations of an Initial Segment Form a Group under Composition. Let $\mathrm{sgn}(\sigma)$ be the sign of $\sigma$; a permutation with no inversion has inversion count $0$, which is read as even, so that its sign is $1$. Powers $c^{k}$ of a real number are those of Natural Number Power of an Element of a Field.

For $p,q\in[n]$ with $p\ne q$, the transposition of $p$ and $q$ is the map $\theta_{pq}:[n]\to[n]$ with

$$\theta_{pq}(p)=q,\qquad \theta_{pq}(q)=p,\qquad \theta_{pq}(k)=k\ \text{ for every }k\in[n]\text{ with }k\ne p\text{ and }k\ne q.$$

Then the following hold.

1. (Values of the sign) $\mathrm{sgn}(\mathrm{id})=1$, and $\mathrm{sgn}(\sigma)\,\mathrm{sgn}(\sigma)=1$ for every $\sigma\in S_{n}$; consequently $\mathrm{sgn}(\sigma)=1$ or $\mathrm{sgn}(\sigma)=-1$.

2. (Multiplicativity) $\mathrm{sgn}(\sigma\circ\tau)=\mathrm{sgn}(\sigma)\,\mathrm{sgn}(\tau)$ for all $\sigma,\tau\in S_{n}$.

3. (Inverses) $\mathrm{sgn}(\sigma^{-1})=\mathrm{sgn}(\sigma)$ for every $\sigma\in S_{n}$.

4. (Transpositions) For all $p,q\in[n]$ with $p\ne q$ the map $\theta_{pq}$ belongs to $S_{n}$ and $\mathrm{sgn}(\theta_{pq})=-1$.