# a problem involving permutation groups

I am given two cycles $(123)$ and $(456)$ and have to find a permutation $\sigma$ (if it exists) such that $\sigma(123) \sigma^{-1} = (456)$.

This is what I tried: let $\sigma(123) \sigma^{-1} = t$ then $t$ is also a 3-cycle, which is equal to $$(\sigma(1)\ \sigma(2) \ \sigma(3))= (456)$$ $$\implies \sigma(1)=4, \sigma(2)= 5, \sigma(3)=6$$ but how do I get complete $\sigma$ from these three expressions? And how I can be sure that such $\sigma$ even exists at the first place?

Any help leading me to the solution will be greatly appreciated, thanks in advance.

• Just extend the $\sigma$ you have obtained to a permutation of $\{1,2,\ldots, 6\}$. For instance, $$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 1 & 2 & 3 \end{pmatrix}$$ works. It doesn't matter what $\sigma$ does on $\{4,5,6\}$ since $\sigma^{-1}$ will undo that in the composition. – Prahlad Vaidyanathan Mar 31 '15 at 9:42
• okay so instead of taking $1,2,3$ as images of $4,5,6$ under $\sigma$, I can take any other permutation of $1,2,3$, right? – Shreya Mar 31 '15 at 9:54

I don't know how general you want this answer to be, but in this specific case note that these two cycles are disjoint and of the same form. So, if $\sigma$ just flips the elements from the active location of one cycle to that of another, then you've got it. So, take $\sigma$ to be $(1\;4)(2\;5)(3\;6)$. Then $\sigma(1\;2\;3)=(4\;5\;6)\sigma$ because each logically rotates the first three elements and then switches them with the last three (the second just performing the rotation after switching between the first three and last three).