Two questions about cycles (algebra)...


by Artusartos
Tags: algebra, cycles
Artusartos
Artusartos is offline
#1
Dec25-12, 11:21 AM
P: 245
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
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Stephen Tashi
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#2
Dec25-12, 04:07 PM
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Quote Quote by Artusartos View Post
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 )
algebrat
algebrat is offline
#3
Dec26-12, 04:23 AM
P: 428
In a nutshell, notice that alpha gamma alpha inverse takes alpha of 1 to alpha of 3. ;)

Artusartos
Artusartos is offline
#4
Dec26-12, 04:39 PM
P: 245

Two questions about cycles (algebra)...


Quote Quote by Stephen Tashi View Post
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


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