If \(\sigma = (i_1 i_2 \cdots i_r) \in S_n\) and \(\tau \in S_n\), then show that \(\tau\sigma\tau^{-1} = (\tau(i_1) \tau(i_2) \cdots \tau(i_r))\).
Solution
Set \[ \rho=(\tau(i_1)\ \tau(i_2)\ \cdots\ \tau(i_r)). \] We prove that \(\tau\sigma\tau^{-1}\) and \(\rho\) agree on every element of \(\{1,\ldots,n\}\).
1. For \(1\le j<r\),\[ (\tau\sigma\tau^{-1})(\tau(i_j))=\tau(i_{j+1}), \]and\[ (\tau\sigma\tau^{-1})(\tau(i_r))=\tau(i_1). \]
For \(j<r\), \[ (\tau\sigma\tau^{-1})(\tau(i_j)) =\tau(\sigma(i_j)) =\tau(i_{j+1}), \] while \[ (\tau\sigma\tau^{-1})(\tau(i_r)) =\tau(\sigma(i_r)) =\tau(i_1). \] These are exactly the values of \(\rho\) on its support.
2. Every point outside \(\{\tau(i_1),\ldots,\tau(i_r)\}\) is fixed by\(\tau\sigma\tau^{-1}\).
Let \(x\notin\{\tau(i_1),\ldots,\tau(i_r)\}\). Since \(\tau\) is bijective, \(\tau^{-1}(x)\notin\{i_1,\ldots,i_r\}\), so \(\sigma\) fixes \(\tau^{-1}(x)\). Therefore \[ (\tau\sigma\tau^{-1})(x) =\tau(\tau^{-1}(x)) =x. \] The cycle \(\rho\) fixes the same points.
3. Hence\[ \tau\sigma\tau^{-1} =(\tau(i_1)\ \tau(i_2)\ \cdots\ \tau(i_r)). \]
By 1 and 2 the two permutations agree on every point.