Transposition is of Odd Parity

Theorem
Let $$S_n$$ denote the set of permutations on $n$ letters.

Let $$\pi \in S_n$$ be a transposition.

Then $$\pi$$ is of odd parity.

Proof
Let $$\pi = \begin{bmatrix} 1 & 2 \end{bmatrix}$$ be a transposition.

Let $$\Delta_n$$ be defined as Product of Differences.

Then $$\forall n \in \N^*: \pi \cdot \Delta_n$$ 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.