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.
$\blacksquare$
