Thanks for the very helpful videos. I wanted to let you know that Sylow's name is not pronounced See-low, it is pronounced similarly to Sue-loav (as in loaves).
Thank you very much for these great videos ! Just a small remark: at 18:40, when you say "each of these subgroups contains a nested chain of p-subgroups", we did not yet prove that there is only one chain (if this is even the case) - couldn't there be a lattice of p-subgroups instead of a chain ? Although the proof only shows how to build one such subgroups, couldn't other methods yield distinct p-subgroups ?
hey! in 11:20 - I wanted to ask in what video do you prove that the order or the quotient group - normalizer of H mod H is a multiple of p? hanks a lot for this video!
Hi, To be fully rigorous, wouldn't you also want to show that H' is a subgroup in your proof od the first Sylow Theorem? If this is obvious, how do you see it?
H' is constructed as the preimage of under the quotient. Then use that the quotient is a group homomorphism and the preimage of a subgroup must be a subgroup.
Does the reasoning behind the normality of P5 generalize? That is, if G contains a subgroup H that is not isomorphic to any other subgroup of G can we conclude that H is normal?
@@ProfessorMacauley Thanks. And also thanks for the great videos. The explanations are clear as crystal. Only a few videos about Advanced Linear Algebra? I am waiting for more. And what about Lie Groups and Representation Theory, two other very interesting topics?
This is by far the best lecture on Sylow theorems one can find on youtube.
This video is by far the clearest group theory lecture I watched. It was so good that some ideas really clicked with me for the first time.
Thank you so much professor. I was stuck on this for last three days
Army
I was stuck for the last 7 years. Ha!
Thank you Professor for these amazing lectures
Thankyou so much for these lectures !!! The way you teach with intuition & understanding is of great help ☺
Amazing videos. Thank you, professor.
Thank you very much sir! I am currently reviewing about Sylow Theorems and this video helped a lot.
in 31:05 it's easy to prove that conjugation of a p-subgroup is also a p-subgroup, so the only orbit is not bigger than np but equal to np
Thank you for such an amazing presentation of this powerful result!
Good pictures, clear explanation--very nicely done.
oh my gosh ,why you so great ! I love you!!!You are the one who teach it really easy.I can't agree more!🥳🥳🥳
Thank you so much. This video was very helpful. I wish you were my lecturer
I believe there's a small notational error at 25:06. I think it should be [G:H] instead of [G,H].
Merci beaucoup et super travail.
Thanks for the very helpful videos. I wanted to let you know that Sylow's name is not pronounced See-low, it is pronounced similarly to Sue-loav (as in loaves).
Thank you very much for these great videos ! Just a small remark: at 18:40, when you say "each of these subgroups contains a nested chain of p-subgroups", we did not yet prove that there is only one chain (if this is even the case) - couldn't there be a lattice of p-subgroups instead of a chain ? Although the proof only shows how to build one such subgroups, couldn't other methods yield distinct p-subgroups ?
Really Helpful with graphs!Thank a lot!
Nice Sylow theorems explanation. Have we the proof of the such normalize group Ng(H) existence ?
hey!
in 11:20 - I wanted to ask in what video do you prove that the order or the quotient group - normalizer of H mod H is a multiple of p?
hanks a lot for this video!
The previous one, on p-groups
Hi,
To be fully rigorous, wouldn't you also want to show that H' is a subgroup in your proof od the first Sylow Theorem? If this is obvious, how do you see it?
H' is constructed as the preimage of under the quotient. Then use that the quotient is a group homomorphism and the preimage of a subgroup must be a subgroup.
It's really very helpful sir...thank you
The best!
amazing
Does the reasoning behind the normality of P5 generalize? That is, if G contains a subgroup H that is not isomorphic to any other subgroup of G can we conclude that H is normal?
Yes! Because in that case, xHx^{-1} must be equal to H, for any x.
@@ProfessorMacauley Thanks. And also thanks for the great videos. The explanations are clear as crystal. Only a few videos about Advanced Linear Algebra? I am waiting for more. And what about Lie Groups and Representation Theory, two other very interesting topics?
Pretty amazing stuff
thank you
wow!, very awesome.
Mmmmm❤
/ˈsyːlɔv/ 😎