Parity Function is Homomorphism

From ProofWiki
Jump to navigation Jump to search

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.

\(\displaystyle \map \sgn {\pi \rho} \Delta_n\) \(=\) \(\displaystyle \pi \rho \cdot \Delta_n\) Definition of Sign of Permutation
\(\displaystyle \) \(=\) \(\displaystyle \pi \cdot \paren {\rho \cdot \Delta_n}\) Permutation on Polynomial is Group Action
\(\displaystyle \) \(=\) \(\displaystyle \pi \cdot \paren {\map \sgn \rho \cdot \Delta_n}\) Definition of Sign of Permutation
\(\displaystyle \) \(=\) \(\displaystyle \pi \, \map \sgn \rho \cdot \Delta_n\) Permutation on Polynomial is Group Action
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn \pi \, \map \sgn \rho \Delta_n\) Definition of Sign of Permutation


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

$\blacksquare$


Sources