K-Cycle can be Factored into Transpositions

Theorem
Every $k$-cycle can be factorised into the product of $k-1$ transpositions.

This factorisation is not unique.

Proof

 * The cycle $\begin{bmatrix} 1 & 2 & \ldots & k \end{bmatrix}$ has the factorisation:

$\begin{bmatrix} 1 & 2 & \ldots & k \end{bmatrix} = \begin{bmatrix} 1 & k \end{bmatrix} \ldots \begin{bmatrix} 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} $

Therefore, the general $k$-cycle $\begin{bmatrix} i_1 & i_2 & \ldots & i_k \end{bmatrix}$ has the factorisation:

$\begin{bmatrix} i_1 & i_2 & \ldots & i_k \end{bmatrix} = \begin{bmatrix} i_1 & i_k \end{bmatrix} \ldots \begin{bmatrix} i_1 & i_3 \end{bmatrix} \begin{bmatrix} i_1 & i_2 \end{bmatrix}$


 * The cycle $\begin{bmatrix} 1 & 2 & \ldots & k \end{bmatrix}$ also has the factorisation:

$\begin{bmatrix} 1 & 2 & \ldots & k \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \end{bmatrix} \ldots \begin{bmatrix} k-1 & k \end{bmatrix}$

Therefore, the general$k$-cycle $\begin{bmatrix} i_1 & i_2 & \ldots & i_k \end{bmatrix}$ also has the factorisation:

$\begin{bmatrix} i_1 & i_2 & \ldots & i_k \end{bmatrix} = \begin{bmatrix} i_1 & i_2 \end{bmatrix} \begin{bmatrix} i_2 & i_3 \end{bmatrix} \ldots \begin{bmatrix} i_{k-1} & i_k \end{bmatrix}$