Exercise 4.3.10 (Find $\tau$ such that $\tau \sigma \tau^{-1} = \sigma^k$)

Proof. $\sigma =(12345)$.

(a)

${\sigma }^2=(13524)$. Put $$\tau =\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 5\hfill & \hfill 2\hfill & \hfill 4\hfill \end{array}\right)=(2354).$$

Then by Proposition 10,

$$\mathit{𝜏𝜎}{\tau }^{-1}=(13524)={\sigma }^2.$$

(b)

${\sigma }^{-1}=(15432)$. Put $$\tau =\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill 4\hfill & \hfill 3\hfill & \hfill 2\hfill \end{array}\right)=(25)(34).$$

Then

$$\mathit{𝜏𝜎}{\tau }^{-1}=(15432)={\sigma }^{-1}.$$

(c)

${\sigma }^{-2}=(14253)$. Put $$\tau =\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 2\hfill & \hfill 5\hfill & \hfill 3\hfill \end{array}\right)=(2453).$$

Then

$$\mathit{𝜏𝜎}{\tau }^{-1}=(14253)={\sigma }^{-2}.$$