Me, myself and conjugate permutations

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Discussion Overview

The discussion revolves around finding a permutation \(\tau\) in \(S_n\) such that \(\rho = \tau^{-1} \sigma \tau\) for given permutations \(\sigma\) and \(\rho\). Participants explore methods for achieving this, particularly in the context of more complex permutations beyond single k-cycles.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Chen inquires about a general method for finding \(\tau\) given \(\sigma\) and \(\rho\), specifically mentioning the complexity of permutations beyond single k-cycles.
  • One participant suggests that the conjugation lemma might be applicable, questioning if it only works with k-cycles.
  • Another participant proposes considering a change of basis in a vector space to realize the permutations as permutations of basis elements.
  • Chen describes a trial-and-error method to find \(\tau\) and provides a specific example, detailing the steps taken to derive \(\tau\) from \(\sigma\) and \(\rho\).
  • One participant expresses confidence that the method will work consistently based on their understanding from a modern algebra course.

Areas of Agreement / Disagreement

Participants express varying levels of confidence in the methods discussed, with some uncertainty about the applicability of the conjugation lemma and the generalizability of Chen's approach. No consensus is reached on a definitive method for all cases.

Contextual Notes

Some participants reference specific mathematical concepts and terminology that may not be universally understood, indicating potential gaps in shared knowledge or assumptions about familiarity with the subject matter.

Chen
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Hi,

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

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

Thanks,
Chen
 
Last edited:
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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 [tex]\tau \in S_6[/tex] such that:
[tex]\tau (1, 2)(3, 4) \tau ^{-1} = (5, 6)(1, 3)[/tex]
Which can be written as:
[tex]\tau (1, 2)\tau ^{-1} \tau (3, 4) \tau ^{-1} = (5, 6)(1, 3)[/tex]
So I assumed that:
[tex]\tau (1, 2) \tau ^{-1} = (5, 6)[/tex]
[tex]\tau (3, 4) \tau ^{-1} = (1, 3)[/tex]
Which means that:
[tex]\tau (1) = 5[/tex]
[tex]\tau (2) = 6[/tex]
[tex]\tau (3) = 1[/tex]
[tex]\tau (4) = 3[/tex]
So I get:
[tex]\tau = (4, 3, 1)(1, 5)(2, 6)[/tex]

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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