# Solving for a Permutation?

The question I'm asked to solve is as follows.

Give an example of the following with a brief justification:

A permutation $$\pi \in S_4$$ such that $(243) =\pi(123)\pi^{-1}$$So far, I know that I can write $$\pi$$ as $$\pi = \left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & b & c & d\end{array}\right]$$ and I can rewrite the above equation as $$(2,4,3)=\pi(1,2,3)\pi^{-1} \rightarrow (2,4,3)\pi = \pi(1,2,3)$$ From there I think I can write the left side of the equation as $$(2,4,3)\pi = (2,4,3)\left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & b & c & d\end{array}\right] = \left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & d & b & c\end{array}\right]$$ but from here I'm unsure where to go. What should I do next? • I think it's$(14)(23)\$ (which is self-inverse) but I can't explain how I found it. Maybe you can try assuming it's a pair of 2-cycles and see if it works. Apr 9, 2021 at 22:37

A very useful fact is that for any cycle $$(a_1a_2a_3\cdots)$$, we have $$\pi(a_1a_2a_3\cdots)\pi^{-1}=(\pi(a_1)\pi(a_2)\pi(a_3)\cdots)$$ So, we need to solve $$(243)=\pi(123)\pi^{-1}=(\pi(1)\pi(2)\pi(3))$$. We can therefore choose $$\pi=(124)$$ as a solution by lining up the indices, i.e. saying $$\pi$$ sends $$(\color{blue}{123})$$ componentwise to $$(\color{red}{243})$$: so $$\color{blue}1\mapsto \color{red}2$$, $$\color{blue}2 \mapsto \color{red}4$$, $$\color{blue}3 \mapsto \color{red}3$$. In matrix form, this would be $$\pi=\left[\begin{matrix}1 &2 &3 &4\\ 2 &4 &3 &(?)\end{matrix}\right]$$, and there's exactly one way to fill in the $$(?)$$.
But this isn't the only solution! Cycle notation is overspecified, so e.g. $$(243) = (432) = (324)$$. We could have gotten the solutions $$\pi = (14)(23)$$, $$\pi = (134)$$ from lining up these other two representations with $$(\pi(1)\pi(2)\pi(3))$$ and following the same procedure.
Note also that if we were dealing with, say $$S_5$$, we wouldn't have been able to uniquely determine where $$4$$ should have gone—it could just as well have gone to $$5$$, or something else if we had more indices to play with!