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:

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