A conjugate of two permutations question

[Ryker]

Homework Statement

Suppose $x_{1} = \begin{pmatrix}<br /> 2 & 9 & 6 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 3 & 5 & 8 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 4 & 7 \\<br /> \end{pmatrix}$ and $x_{2} = \begin{pmatrix}<br /> 1 & 5 & 9 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 2 & 7 & 6 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 3 & 4 \\<br /> \end{pmatrix}.$

Determine the conjugate a, so that x₁ = ax₂a^-1.

The Attempt at a Solution

I know the solution is a = (1 6 8)(2 3 7 5), since we did this in class. However, we didn't really explain how we got to this solution. And I can do conjugates where you just line up the cycles one under the other, but this method doesn't work here, because, say 1 does not get sent to 2, and 5 not to 9, as you'd assume if you just wrote (1 5 9) above (2 9 6).

I really want to figure this out, but this example really puzzles me, as I haven't yet found the general method, and the fact that x_2,i = a(x_1,i) doesn't really help me here.

Anyways, any help here would be greatly appreciated.

 

[HallsofIvy]

I'm not sure what you mean by "line up the cycles" but I think you meant writing $x_1$ as
$$\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\1 & 6 & 5 & 7 & 3 & 9 & 4 & 5 & 2\end{pmatrix}$$
and $x_2$ as
$$\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\9 & 6 & 4 & 3 & 1 & 7 & 2 & 8 & 5\end{pmatrix}$$

Do you see how I got that? (I am assuming you are using the convention that your permutations work from right to left.)

 

[Ryker]

Yeah, I can follow that, but I can't seem to be able to make the next step then. For example, it isn't obvious to me why a sends 1 to 6 instead of say, to 2 or 9.