Conjugates of Transpositions

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Theorem

Let $k_1, k_2, k_3 \in \left\{{1, 2, \ldots, n}\right\}$.

Then:

$(1): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_3 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix}$

$(2): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_1 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_3 & k_1 \end{bmatrix}$

Proof

$(1):$ Calculating the product of $\begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_3 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix}$:

$k_1 \to k_3 \to k_2$

$k_2 \to k_3 \to k_1$

$k_3 \to k_2 \to k_3$

hence the result.

$\Box$

$(2):$ Calculating the product of $\begin{bmatrix} k_3 & k_1 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_3 & k_1 \end{bmatrix}$:

$k_1 \to k_3 \to k_2$

$k_2 \to k_3 \to k_1$

$k_3 \to k_1 \to k_3$

hence the result.

$\blacksquare$

Conjugates of Transpositions

Theorem

Proof

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