Me, myself and conjugate permutations

Chen
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Hi,

Is there a general method, given \sigma and \rho in Sn, for finding a permutation \tau in Sn such that \rho = \tau ^{-1} \sigma \tau? I know how to do it when \sigma and \rho are made of a single k-cycle, but what happens when they are more complex?

For example, for:
\sigma = (1, 2)(3, 4)
\rho = (5, 6)(1, 3)
In S6.

Thanks,
Chen
 
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Do you know you've just solved me a problem I've been working on all day? (By reminding me of conjugation).

Now to current business: I don't remember it precisely, but isn't the conjugation lemma suppose to do the trick? Or does it work only with k-circles?
 
As a hint you could try thinking of change of basis in a vector space where we realize the permutations as a permutation of basis elements. That ought to work.
 
Palindrom said:
Do you know you've just solved me a problem I've been working on all day? (By reminding me of conjugation).

Now to current business: I don't remember it precisely, but isn't the conjugation lemma suppose to do the trick? Or does it work only with k-circles?
Glad I could help. :biggrin:

I figured it out, by trial and error. I needed to find \tau \in S_6 such that:
\tau (1, 2)(3, 4) \tau ^{-1} = (5, 6)(1, 3)
Which can be written as:
\tau (1, 2)\tau ^{-1} \tau (3, 4) \tau ^{-1} = (5, 6)(1, 3)
So I assumed that:
\tau (1, 2) \tau ^{-1} = (5, 6)
\tau (3, 4) \tau ^{-1} = (1, 3)
Which means that:
\tau (1) = 5
\tau (2) = 6
\tau (3) = 1
\tau (4) = 3
So I get:
\tau = (4, 3, 1)(1, 5)(2, 6)

Hopefully though this wasn't a fluke and this method will work all the time. Matt, unfortunately I don't really know what you're talking about... :blushing: or maybe I know it by a different name. Thanks thought.
 
It should work every time, if I remember my first modern algebra course correctly.

Oh, and Hag Sameah :)
 
Thank you very much, Happy Passover. :smile:
 
Pessah my friend, I'm from Haifa.
 
Yeah, I thought so. I think I know you from ASAT. :wink:
 
It's a small world after all... :smile:
 
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