Parity Function is Homomorphism

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

Let $\pi \in S_n$.

Let $\operatorname{sgn} \left({\pi}\right)$ be the sign of $\pi$.

The parity function of $\pi$ is defined as:


 * Parity of $\pi = \begin{cases}

\mathrm {Even} & : \operatorname{sgn} \left({\pi}\right) = 1 \\ \mathrm {Odd} & : \operatorname{sgn} \left({\pi}\right) = -1 \end{cases}$

The mapping $\operatorname{sgn}: S_n \to C_2$, where $C_2$ is the cyclic group of order 2, is a homomorphism.

Proof
We need to show that $\forall \pi, \rho \in S_n: \operatorname{sgn} \left({\pi}\right) \operatorname{sgn} \left({\rho}\right) = \operatorname{sgn} \left({\pi \rho}\right)$.

Let $\Delta_n$ be an arbitrary product of differences.

As $\left({\left\{{1, -1}\right\}, \times}\right)$ is the parity group, the result follows immediately.