Parity of K-Cycle

Theorem
Let $$\pi$$ be a $k$-cycle.

Then $$\sgn {\pi} = \begin{cases} 1 & : k \ \mathrm {odd} \\ -1 & : k \ \mathrm {even} \end{cases}$$.

Proof

 * First we show that a transposition is of odd parity.

Let $$\pi = \begin{bmatrix} 1 & 2 \end{bmatrix}$$.

Then $$\forall n \in \mathbb{N}^*: \pi \cdot \Delta_n$$ produces only one sign change in $$\Delta_n$$, that is, the one occurring in the factor $$\left({x_1 - x_2}\right)$$.

Thus $$\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \Delta_n = - \Delta_n$$, and thus $$\begin{bmatrix} 1 & 2 \end{bmatrix}$$ is odd.


 * From Conjugates of Transpositions, $$\begin{bmatrix} 1 & k \end{bmatrix} = \begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}$$.

Thus as $\begin{bmatrix} 2 & k \end{bmatrix}$ is self-inverse, $$\begin{bmatrix} 1 & k \end{bmatrix} = \begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}^{-1}$$.

But from Parity of Conjugate of Permutation, $$\sgn \left({\begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}^{-1}}\right) = \sgn \left({\begin{bmatrix} 1 & 2 \end{bmatrix}}\right)$$.

Thus $$\sgn \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right) = \sgn \left({\begin{bmatrix} 1 & 2 \end{bmatrix}}\right)$$ and thus $$\begin{bmatrix} 1 & k \end{bmatrix}$$ is odd.


 * Finally:

$$ $$

But from Parity of Conjugate of Permutation, $$\sgn \left({\begin{bmatrix} 1 & h \end{bmatrix} \begin{bmatrix} 1 & k \end{bmatrix} \begin{bmatrix} 1 & h \end{bmatrix}^{-1}}\right) = \sgn \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right)$$.

Thus $$\sgn \left({\begin{bmatrix} h & k \end{bmatrix}}\right) = \sgn \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right)$$ and thus $$\begin{bmatrix} h & k \end{bmatrix}$$ is odd.


 * Finally, by A K-Cycle can be Factored into Transpositions, we see that a $k$-cycle is the product of $$k-1$$ transpositions, and thus is even iff $$k$$ is odd.