Two questions about cycles (algebra)

  • Context: Undergrad 
  • Thread starter Thread starter Artusartos
  • Start date Start date
  • Tags Tags
    Algebra Cycles
Click For Summary

Discussion Overview

The discussion revolves around two algebraic questions related to cycles in permutation groups, specifically focusing on the application of a theorem regarding the manipulation of cycles and the computation of certain values in a table related to these cycles.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question the expression \(\alpha\gamma\alpha^{-1}\) and why it is represented as \((\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6)\) instead of including inverses of the symbols in the cycles.
  • It is noted that the theorem states to "apply \(\alpha\)" to the symbols in the cycles, suggesting that the cycle structure is significant and not merely a product of individual symbols.
  • One participant emphasizes that a cycle, such as \((1,2,3)\), is not equivalent to the product of its individual symbols, highlighting a conceptual distinction.
  • Another participant explains that in the permutation group \(S_4\), there are 8 distinct elements that are cycles of length 3, using the example of \((1,2,3)\) to illustrate this point.
  • They further clarify that while there are 24 permutations of 3 distinct numbers from the set \{1,2,3,4\}, each cycle has 3 representations, leading to 8 distinct cycles of length 3.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the theorem and the representation of cycles, indicating that the discussion remains unresolved regarding the correct approach to these algebraic expressions.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the theorem and the definitions of cycles, which are not fully explored or clarified.

Artusartos
Messages
236
Reaction score
0
I have two questions:

1) For the example on the second page, I don't understand why they say [tex]\alpha\gamma\alpha^{-1} = (\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6)[/tex] instead of [tex]\alpha\gamma\alpha^{-1} = (\alpha1\alpha^{-1} \alpha3\alpha^{-1})(\alpha2\alpha^{-1} \alpha4\alpha^{-1} \alpha7\alpha^{-1})(\alpha5\alpha^{-1})(\alpha6\alpha^{-1})[/tex].

2) For the tables at the top of the 2nd page, I don't know how they computed those numbers...

Thanks in advance
 

Attachments

  • 20121225_114029.jpg
    20121225_114029.jpg
    13.8 KB · Views: 431
  • 20121225_114339.jpg
    20121225_114339.jpg
    32.2 KB · Views: 464
Last edited:
Physics news on Phys.org
Artusartos said:
I have two questions:

1) For the example on the second page, I don't understand why they say [tex]\alpha\gamma\alpha^{-1} = (\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6)[/tex] instead of [tex]\alpha\gamma\alpha^{-1} = (\alpha1\alpha^{-1} \alpha3\alpha^{-1})(\alpha2\alpha^{-1} \alpha4\alpha^{-1} \alpha7\alpha^{-1})(\alpha5\alpha^{-1})(\alpha6\alpha^{-1})[/tex].

They say what is consistent with what the theorem says. The theorem says to "apply [itex]\alpha[/itex]" to the symbols in the cycles.

If [itex]\alpha,\ p,\ q[/itex] are cycles, It is true that [itex]\alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1})[/itex] but this is not the content of the theorem. A cycle is not the same as the product of the individual symbols in the cycle. The cycle (1,2,3) is not equal to (1)(2)(3).

2) For the tables at the top of the 2nd page, I don't know how they computed those numbers...

For example, In the permutation group [itex]S_4[/itex], there are 8 different elements of the group that are cycles of length 3. The example (1,2,3) in the table illustrates one of them.
(There are 24 = (4)(3)(2) different permutations that can be formed by taking 3 distinct numbers from the set of numbers {1,2,3,4}. However, each permutation such as (1,2,3) is one of 3 representations of the same cycle. (1,2,3) = (2,3,1) = (3,1,2) So there are 8 = 24/3 distinct cycles of length 3 )
 
In a nutshell, notice that alpha gamma alpha inverse takes alpha of 1 to alpha of 3. ;)
 
Stephen Tashi said:
They say what is consistent with what the theorem says. The theorem says to "apply [itex]\alpha[/itex]" to the symbols in the cycles.

If [itex]\alpha,\ p,\ q[/itex] are cycles, It is true that [itex]\alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1})[/itex] but this is not the content of the theorem. A cycle is not the same as the product of the individual symbols in the cycle. The cycle (1,2,3) is not equal to (1)(2)(3).



For example, In the permutation group [itex]S_4[/itex], there are 8 different elements of the group that are cycles of length 3. The example (1,2,3) in the table illustrates one of them.
(There are 24 = (4)(3)(2) different permutations that can be formed by taking 3 distinct numbers from the set of numbers {1,2,3,4}. However, each permutation such as (1,2,3) is one of 3 representations of the same cycle. (1,2,3) = (2,3,1) = (3,1,2) So there are 8 = 24/3 distinct cycles of length 3 )

Thank you
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K