A conjugate of two permutations question

[Ryker]

Homework Statement

Suppose x_{1} = \begin{pmatrix}<br /> 2 &amp; 9 &amp; 6 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 3 &amp; 5 &amp; 8 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 4 &amp; 7 \\<br /> \end{pmatrix} and x_{2} = \begin{pmatrix}<br /> 1 &amp; 5 &amp; 9 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 2 &amp; 7 &amp; 6 \\<br /> \end{pmatrix}\begin{pmatrix}<br /> 3 &amp; 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 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 \\1 &amp; 6 &amp; 5 &amp; 7 &amp; 3 &amp; 9 &amp; 4 &amp; 5 &amp; 2\end{pmatrix}
and x_2 as
\begin{pmatrix}1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 \\9 &amp; 6 &amp; 4 &amp; 3 &amp; 1 &amp; 7 &amp; 2 &amp; 8 &amp; 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.