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A conjugate of two permutations question

  • Thread starter Ryker
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  • #1
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Homework Statement


Suppose [itex]x_{1} = \begin{pmatrix}
2 & 9 & 6 \\
\end{pmatrix}\begin{pmatrix}
3 & 5 & 8 \\
\end{pmatrix}\begin{pmatrix}
4 & 7 \\
\end{pmatrix}[/itex] and [itex]x_{2} = \begin{pmatrix}
1 & 5 & 9 \\
\end{pmatrix}\begin{pmatrix}
2 & 7 & 6 \\
\end{pmatrix}\begin{pmatrix}
3 & 4 \\
\end{pmatrix}.[/itex]

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 x2,i = a(x1,i) doesn't really help me here.

Anyways, any help here would be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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I'm not sure what you mean by "line up the cycles" but I think you meant writing [itex]x_1[/itex] as
[tex]\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\1 & 6 & 5 & 7 & 3 & 9 & 4 & 5 & 2\end{pmatrix}[/tex]
and [itex]x_2[/itex] as
[tex]\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\9 & 6 & 4 & 3 & 1 & 7 & 2 & 8 & 5\end{pmatrix}[/tex]

Do you see how I got that? (I am assuming you are using the convention that your permutations work from right to left.)
 
Last edited by a moderator:
  • #3
1,086
2
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.
 

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