(cont.) That gives the centralizer in S_n. For A_n, we only keep the even permutations. Example: (123) in A_5. In S_5, we have e, (123), (132), (45), (123)(45), (132)(45). The only even elements are e, (123), (132).
First review the Conjugation Rule in GT11.1. Autos of A4. When we compute rsr^-1, the cycle structure of s is unchanged, but we relabel according to r. If rsr^-1 = s, then the labels for each cycle in s map back to themselves. In fact, they are only allowed to shift to remain unchanged - this happens only by using powers of that cycle in r. On the other hand, the labels unused by s must also map back to themselves. No restrictions here. This contributes a possible S_k.
Hi, at 8:18 I'm not sure if I understand the argument here correctly. I'll state how I understood it explicitly and if you have time please let me know if I have the right idea. We are looking for elements in S_n that don't permute the cycle (123). These are contained in the product (gh) iso to (g,h) where g is in S_3 and h is in S_(n-3). Elements h do not permute (123) because they are disjoint so we get all of S_(n-3)) in the centralizer, plus some elements in S_3. That is I'm assuming that the intersection of S_3 and S_(n-3) only has the trivial element. Now moving into A_n we note that the only elements in A_3 that don't permute (123) under conjugation are those in the s.g (ie Z/3) and these elements coupled with any elements from A_(n-3) by the direct product X A_(n-3). Hence we get that the order of the centralizer of (123) as 3*(n-3)!/2. This seems correct, but I don't have the confidence just yet. :)
I'm sorry, could you explain further why the Centralizers of the elements turn out to be that way? I can see clearly that for example C(123) contains the elements e, (123) and (132)=(123)^{-1}, but why can't there be any others?
(cont.) That gives the centralizer in S_n. For A_n, we only keep the even permutations.
Example: (123) in A_5. In S_5, we have e, (123), (132), (45), (123)(45), (132)(45).
The only even elements are e, (123), (132).
First review the Conjugation Rule in GT11.1. Autos of A4. When we compute rsr^-1, the cycle structure of s is unchanged, but we relabel according to r.
If rsr^-1 = s, then the labels for each cycle in s map back to themselves. In fact, they are only allowed to shift to remain unchanged - this happens only by using powers of that cycle in r.
On the other hand, the labels unused by s must also map back to themselves. No restrictions here. This contributes a possible S_k.
Hi, at 8:18 I'm not sure if I understand the argument here correctly. I'll state how I understood it explicitly and if you have time please let me know if I have the right idea. We are looking for elements in S_n that don't permute the cycle (123). These are contained in the product (gh) iso to (g,h) where g is in S_3 and h is in S_(n-3). Elements h do not permute (123) because they are disjoint so we get all of S_(n-3)) in the centralizer, plus some elements in S_3. That is I'm assuming that the intersection of S_3 and S_(n-3) only has the trivial element.
Now moving into A_n we note that the only elements in A_3 that don't permute (123) under conjugation are those in the s.g (ie Z/3) and these elements coupled with any elements from A_(n-3) by the direct product X A_(n-3).
Hence we get that the order of the centralizer of (123) as 3*(n-3)!/2.
This seems correct, but I don't have the confidence just yet. :)
That's pretty much it.
I'm sorry, could you explain further why the Centralizers of the elements turn out to be that way? I can see clearly that for example C(123) contains the elements e, (123) and (132)=(123)^{-1}, but why can't there be any others?
at 10:29 isn't P the LARGEST prime that divides m?
thanks again for your excellent videos.
Your welcome! I don't think it matters. Any prime divisor should work.
sir please provide subtitles for all ur videos.iam unable to catch ur english.i need to understand all of ur videos.but i am not getting it .
super, thank you very much