This playlist is the most comprehensible presentation of Abstract Algebra that I have found. Somehow the sound has disengaged from this particular video. Either that or I'm having a very bad reaction to something I ate.
At 4:42, (123) and (124) verify the identity (123) = (34)(124)(34) so they can't be conjugate in A_4 (equation (124)=g(123)g^(-1)) because otherwise (124) would commute with g(34), which is odd so there would be only one 3-cycle conjugation class in A_4.
His error was in mixing two kinds of permutations in the Cauchy diagram - one type (which is standard) in which you place the next item in the cycle below the current item, and the other type where you place the number in the *position* (indexed by the original 1234) corresponding to the next item in the cycle. He needs to choose one or the other, not mix both.
So if I understand well. The S5 ad the A5 are not solvable. But the D5 group is? But how is a eqation of the D5 group solvable? For example X^5 + -5*X +12 = 0 and X^5 + 20*X +32 = 0 have group D5. How can you solve them then?
This playlist is the most comprehensible presentation of Abstract Algebra that I have found.
Somehow the sound has disengaged from this particular video. Either that or I'm having a very bad reaction to something I ate.
Yeah, the audio was pretty desynced from what I saw and heard too.
Cycle types? More like "Cool videos where knowledge is ripe!" 👍
At 4:42, (123) and (124) verify the identity
(123) = (34)(124)(34)
so they can't be conjugate in A_4 (equation (124)=g(123)g^(-1)) because otherwise (124) would commute with g(34), which is odd so there would be only one 3-cycle conjugation class in A_4.
Yes (143) should be on the right, but anyway these videos are excellent. Terrific job :)
At 2:35, (134) should be on the left and (143) should be on the right.
(143) is g and (134) is g inverse.
Thank you!!
His error was in mixing two kinds of permutations in the Cauchy diagram - one type (which is standard) in which you place the next item in the cycle below the current item, and the other type where you place the number in the *position* (indexed by the original 1234) corresponding to the next item in the cycle. He needs to choose one or the other, not mix both.
So if I understand well. The S5 ad the A5 are not solvable. But the D5 group is?
But how is a eqation of the D5 group solvable? For example X^5 + -5*X +12 = 0 and X^5 + 20*X +32 = 0 have group D5. How can you solve them then?
4:33 - - - (23)(123)(23) = (132)
But, (23) is not in A4. So, (123) and (132) are not conjugate.