Question

Let$\tau$ be a transposition and let$\sigma \in S_n$. prove that$\sigma \tau {\sigma }^{-1}$is a transposition.

Short answer

It is proved that $\sigma \tau {\sigma }^{-1}$is a transposition.

Step-by-step solution

Consider

Assume that $\tau =\left(ab\right)\in S_n$be a transposition and$\sigma \in S_n$ be a permutation.

Prove the result 

Consider $c\ne a\text{'}b\text{'}$, we can say that:

$\begin{array}{rcl}\sigma \tau {\sigma }^{-1}\sigma \left(a\right)& =& \sigma \tau \left(a\right)\\ & =& \sigma \left(b\right)\\ \sigma \tau {\sigma }^{-1}\sigma \left(b\right)& =& \sigma \tau \left(b\right)\\ & =& \sigma \left(a\right)\\ \sigma \tau {\sigma }^{-1}\left(c\right)& =& \sigma {\sigma }^{-1}\left(c\right)\\ & =& c\\ & & \end{array}$

This implies that$\sigma \tau {\sigma }^{-1}=\left(\sigma \left(a\right)\sigma \left(b\right)\right)$ is a transposition.

Hence, it is proved that $\sigma \tau {\sigma }^{-1}$is a transposition.