Let \(k\ge 1\), \(\Pi_k\) be the set of pairings of \(\{1,\ldots ,2k\}\) and \(a,b\in \{1,\ldots ,2k\}\) with \(a\neq b\). The transposition map \(\tau_{a,b}\) satisfies
\[ \tau_{a,b}\circ \tau_{a,b}=\id. \]
Note
We can use the group action and the identity \(\tau_{a,b}=\phi_{(a~~b)}\) to find
\[ \tau_{a,b} \circ \tau_{a,b}=\phi_{(a~~b)} \circ \phi_{(a~~b)} = \phi_{(a~~b)(a~~b)} = \phi_{\id} = \id. \]
Proof
Let \(\pi\in \Pi_k\).
If \(b=\pi(a)\), then by definition \(\tau_{a,b}(\pi)=\pi\), and clearly \(\tau_{a,b}\circ \tau_{a,b}(\pi)=\pi\).
Assume now that \(b\neq \pi(a)\),
Applying \(\tau_{a,b}\) once swaps the partners of \(a\) and \(b\), and applying it again restore the original partners, hence we recover \(\pi\).