Me, myself and conjugate permutations

In summary, there is a general method for finding a permutation \tau in Sn such that \rho = \tau ^{-1} \sigma \tau. It involves using the conjugation lemma and thinking of change of basis in a vector space. This method was used to find \tau for the given example in S6. The method should work every time and is commonly taught in modern algebra courses. The conversation also includes greetings and small talk between the speakers.
  • #1
Chen
977
1
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
 
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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
It should work every time, if I remember my first modern algebra course correctly.

Oh, and Hag Sameah :)
 
  • #6
Thank you very much, Happy Passover. :smile:
 
  • #7
Pessah my friend, I'm from Haifa.
 
  • #8
Yeah, I thought so. I think I know you from ASAT. :wink:
 
  • #9
It's a small world after all... :smile:
 

FAQ: Me, myself and conjugate permutations

What is a conjugate permutation?

A conjugate permutation is a rearrangement of the elements in a permutation, where the order of the elements remains the same but the elements themselves are replaced with their corresponding conjugates.

What is the significance of conjugate permutations?

Conjugate permutations are important in group theory and algebra, as they can provide insight into the structure and symmetry of groups. They also have applications in combinatorics and cryptography.

How do you find the conjugate of a permutation?

To find the conjugate of a permutation, you must first identify the corresponding conjugates of each element in the permutation. Then, you can rearrange the elements in the same order as the original permutation, but with their conjugates substituted in.

Can a permutation be conjugate to itself?

Yes, a permutation can be conjugate to itself if all of its elements are self-conjugate. This means that the conjugate of each element is equal to the element itself.

What is the difference between a conjugate permutation and an inverse permutation?

A conjugate permutation is a rearrangement of the elements in a permutation using their conjugates, while an inverse permutation is a rearrangement of the elements in reverse order. In other words, a conjugate permutation preserves the order of the elements, while an inverse permutation reverses the order.

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