Parity Function is Homomorphism

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

Let $\pi \in S_n$.

Let $\map \sgn \pi$ be the sign of $\pi$.

Let the parity function of $\pi$ be defined as:


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

\mathrm {Even} & : \map \sgn \pi = 1 \\ \mathrm {Odd} & : \map \sgn \pi = -1 \end{cases}$

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

Proof
We need to show that:
 * $\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$

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

As $\struct {\set {1, -1}, \times}$ is the parity group, the result follows immediately.