Find the product of the permutations $\sigma\tau\sigma^{-1}$ where $\sigma=(1 6 8)(2 5 4 3)$, $\tau=(1 4 8 7)(2 3)(5 6).$

Question

Are my answers correct?

$$\sigma^{-1}= (8 6 1)(3 4 5 2)$$

$$\implies \sigma\tau\sigma^{-1}= (1 6 8)(2 5 4 3)(1 4 8 7)(2 3)(5 6)(8 6 1)(3 4 5 2)$$

$$\implies \sigma\tau\sigma^{-1}= (1 7 6 3)(2 5)(4 8);$$

also, $|\sigma\tau\sigma^{-1}| = {\rm lcm}(4,2,2) = 4.$

Answer 1

Your answers are correct; although it might help to format your composition in a more understandable way. $\sigma, \tau$ and $\sigma^{-1}$ are functions (bijections), and so $\sigma\tau\sigma^{-1}$ should really be $\sigma\circ\tau\circ\sigma^{-1}$, which contains the composition operator. That would make the second line in your answer more understandable and technically correct: $$\sigma\circ\tau\circ\sigma^{-1} = (168)(2543)\circ(1487)(23)(56)\circ(861)(3452).$$

Answer 2

Conjugation of permutations follows a neat rule, a proof of which is a pleasant exercise and the statement can be found in any algebra textbook worth its salt. Here it is:

Theorem: Let $\rho, \pi\in S_n$ such that $\pi=(p_{1}\dots p_{i_1})(p_2\dots p_{i_2})\dots (p_m\dots p_{i_m})$. Then $$\rho\pi\rho^{-1}=(\rho(p_{1})\dots \rho(p_{i_1}))(\rho(p_2)\dots \rho(p_{i_2}))\dots (\rho(p_m)\dots \rho(p_{i_m})),$$ where $\rho(x)$ is $\rho$ evaluated at $x$.

Let's apply the theorem here. We have

$$\begin{align} \sigma\color{red}{\tau}\sigma^{-1}&=\color{red}{(}(1 6 8)(2 5 4 3)\color{red}{1}, (1 6 8)(2 5 4 3)\color{red}{4}, (1 6 8)(2 5 4 3)\color{red}{8}, (1 6 8)(2 5 4 3)\color{red}{7}\color{red}{)(}(1 6 8)(2 5 4 3)\color{red}{2}, (1 6 8)(2 5 4 3)\color{red}{3}\color{red}{)(}(1 6 8)(2 5 4 3)\color{red}{5}, (1 6 8)(2 5 4 3)\color{red}{6}\color{red}{)}\\ &=(6317)(52)(48), \end{align}$$

which is equivalent to your answer.