Generating Set for the Symmetric Group - Question

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Discussion Overview

The discussion revolves around the generation of the symmetric group Sn using specific permutations, particularly the transposition (1 2) and the cycle (1 2 ... n). Participants explore the implications of these permutations and seek to understand how they can generate all elements of Sn, including transpositions and their relationships.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that (1 2 ... n) can be expressed in terms of transpositions, specifically (1 2) (1 3) ... (1 n), but is unsure how this leads to generating all of Sn.
  • Another participant proposes experimenting with small values of n to gain insights into the generation of Sn.
  • A participant shares specific transpositions they derived for n = 5, but expresses uncertainty about how to generalize this for larger n and about the validity of their approach.
  • Some participants notice a potential pattern in the derived expressions and question whether it holds for other values of n, such as 4 or 6.
  • One participant suggests conjugating (1 2) by (1 ... n) and its powers as a method to explore the generation of transpositions.
  • Another participant confirms that the inverse of (1 ... n) is a power of (1 ... n) and discusses the implications of using inverses in this context.
  • A participant expresses confusion about the process of conjugation and seeks clarification on how to perform it.
  • One participant explains the conjugation process, indicating that it involves calculating the product of the inverse of one permutation with another permutation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the methods for generating all transpositions in Sn or the completeness of the derived expressions. Multiple competing views and approaches remain, with ongoing exploration of patterns and techniques.

Contextual Notes

Participants express uncertainty regarding the generalization of their findings and the completeness of their derived sets of transpositions. There are also limitations in the understanding of conjugation and its application to the problem.

MattL
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Explain why the permutations (1 2) and (1 2 ... n) generate all of Sn, the symmetric group (the group of all permutations of the numbers {1,2,...,n}?

Perhaps something to do with the fact that
(1 2 ... n) = (1 2) (1 3) ... (1 n)?
Other than that I haven't got a clue - help! (please!)

Thanks
 
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Try working it out directly for small n; that might give you ideas.
 
After giving it a go, I've managed to get, for n = 5

(2 3) = (1 2 3 4 5)^4 (1 2) (1 2 3 4 5)
(3 4) = (1 2 3 4 5)^3 (1 2) (1 2 3 4 5)^2
(4 5) = (1 2 3 4 5)^2 (1 2) (1 2 3 4 5)^3


So I can get the transpositions (1 2), (2 3), (4 5), but I can't work out how to get all of the others that I am missing, these being (1 3), (1 4), (1 5), (2 4), (2 5), (3 5). Even if I could, I'm not sure I'd know how to generalise for n. Also, I'm not entirely sure why the compositions I've used above work or where the pattern comes from - I used my computer to do some guesswork for those.

ps if i don't reply, it's because I've gone to bed!
 
Last edited:
Doesn't there seem to be a pattern there? Does it apply for other n? Say, 4 or 6?
 
why don't you conjugate (12) by (1,..,n) and powers of (1,..,n)?
 
Hurkyl said:
Doesn't there seem to be a pattern there? Does it apply for other n? Say, 4 or 6?

Yep, there seems to be a pattern of
(m+1 m+2) = (1 2 ... n)^(n-m+1) (1 2) (1 2 ... n)^(m-1)
and after trying some in Maple, I think that it does hold for other n.
Still, that doesn't give me a complete set of transpositions.

(Another thing - would be ok to use the inverse of (1 2 ... n) in my answer? Not that I know how I would, but does the original question allow this?)
 
The inverse of (1,..,n) is a power of (1,..,n) (but usually one allows inverses as well, though for a finite group this is not explicitly needed - do you see why?).

What makes you think you should get a list of all transpositions from the expressions you've written down? You may then compose (indeed conjugate) the transpositions (m, m+1) and (m+1,m+2) and still be in the set generated by (12) and (1..n)
 
matt grime said:
why don't you conjugate (12) by (1,..,n) and powers of (1,..,n)?

how do I conjugate them? I've only recently started permutations and group theory (and yet I've been given this question :confused: ), so my knowledge is still quite basic ( :frown: )

EDIT: oh, I see...
 
To conjugate g by h one works out the product h**(-1)gh, where h**(-1) is h inverse. My caret key is broken on my laptop.
 

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