24:56 I don't quite see the direct connection why should the other orbits say O_i has to have cardinality >=2? Previously we've got that H= H_i. So, if H is the only Singleton H-orbit in S then via the orbit decomposition we should have got the cardinality of other orbits except O_i is 1, which is same as O_1. Isn't it? Or I guess I'm missing something in between? Could anyone please address this! Thank you!
Let say H2€S then orbit of H2 are hH2h inverse for all h€H.There are p^e element in H and H2. so there are chances that orbit of H2 captain p^e elements(subgroup).one of the element for sure will be H2 because H contains identity element (e) so eH2e inverse result in H2. From here there are two case Case 1: all element of orbit of H2 are H2 in that case sir proved that H2=H Case 2: at least one of the element of orbit H2 is different from H2 .so orbit of H2 have at least two elements. So the orbit of element (subgroup) of set S which is different from H will have at least 2 element
I think your statement just proves your question. Basically if you shoe that for any Hi if its orbit has one element then it means that Hi must be H. Since this forces any Hi element that has single cycle to be H then the rest should have more than one cycle. Because one you assume hHih-1=Hi for same Hi for all h in H then you showed that Hi is in fact H due to H being part of normalizers of any Hi therefore applying Sylow on Ni since any normalizer is subgroup and both H and Hi are in there there fore any the g that conjugates them by Sylow on Ni then that g is also in Ni by definition of congugarion of Hi in Ni. H
@24:00 should read "so H is the only singleton H-orbit in S"
Great session
24:56 I don't quite see the direct connection why should the other orbits say O_i has to have cardinality >=2? Previously we've got that H= H_i. So, if H is the only Singleton H-orbit in S then via the orbit decomposition we should have got the cardinality of other orbits except O_i is 1, which is same as O_1. Isn't it? Or I guess I'm missing something in between? Could anyone please address this! Thank you!
Let say H2€S then orbit of H2 are hH2h inverse for all h€H.There are p^e element in H and H2. so there are chances that orbit of H2 captain p^e elements(subgroup).one of the element for sure will be H2 because H contains identity element (e) so eH2e inverse result in H2. From here there are two case
Case 1: all element of orbit of H2 are H2 in that case sir proved that H2=H
Case 2: at least one of the element of orbit H2 is different from H2 .so orbit of H2 have at least two elements.
So the orbit of element (subgroup) of set S which is different from H will have at least 2 element
I think your statement just proves your question. Basically if you shoe that for any Hi if its orbit has one element then it means that Hi must be H. Since this forces any Hi element that has single cycle to be H then the rest should have more than one cycle. Because one you assume hHih-1=Hi for same Hi for all h in H then you showed that Hi is in fact H due to H being part of normalizers of any Hi therefore applying Sylow on Ni since any normalizer is subgroup and both H and Hi are in there there fore any the g that conjugates them by Sylow on Ni then that g is also in Ni by definition of congugarion of Hi in Ni. H
Beautiful
Thank you so much sir, 🙏
🙏🙏🙏🙏 Thank u very much sir.
Thank you sir
why is N(i) subgroup of N ?? Can u pls explain ??
I think N was supposed to be G. G works, since Ni is taking elements from G, and therefore a subset of G.