Parity Function is Homomorphism

Theorem
$\newcommand{\sgn}{\operatorname{sgn}}$ Let $$S_n$$ denote the symmetric group on $n$ letters.

Let $$\pi \in S_n$$.

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

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


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

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

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

Proof

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

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.