Most classes don't get to it. And many people who do learn it tend to have no idea what's going on cause it's usually super unclear! Thanks for watching Eleazar!
It is the way to classify small finite groups. All the homework after is like "show what possible groups of order 75 up to isomorphism" and then your factor and apply the theorems.
Honestly, before you even started talking about the Theorem, I subscribed. Great video and and thanks for showing regular examples, I needed to understand this for my final.
I miss this stuff. I took Abstract Algebra 1987-1988 to prepare for graduate school and then again 1988 to 1989 in graduate school. So it's been 34 years. I definitely need a refresher.
Note that Omega is a disjoint union of its orbits, this implies that the size of Omega is the sum of the sizes of each orbit. Since Omega is finite, there are certainly a finite number of orbits and the size of each orbit is finite. If O_1,...,O_r are all the orbits of the action, then we can express that fact as |Omega|=|O_1|+...+|O_r|. Suppose every orbit O has size a multiple of p. Then we can write for each i=1,...,r the following: |O_i|=p·h_i for some natural number h_i. Thus, we have that |Omega|=ph_1+...+ph_r=p(h_1+...+h_r), meaning that |Omega| is a multiple of p. But this is a contradiction because we have established by the combinatorial lemma that |Omega| is congruent to m modulo p, and by hypothesis, gcd(m,p)=1, so definitely not a multiple of p.
Wonderful! First time I getting it. I like the technique of keeping relevant information on the side of the board. I'm looking forward to The Two Towers and The Return of the King!
Thanks for the clear explanation. And by the way,of all other proofs of Sylow, this one really stands out for we never need to appeal to the class equation!
I’m preparing for the algebra qual and this was pretty helpful! A much simpler proof than the standard inductive proof starting with cauchys theorem as a base case (although that gives subgroups of size p^j for all j
Thanks for making this easy to understand. Please do more videos on group theory.
Definitely and thanks for watching
Sir, thank you for this video! Greetings from Kygyzstan! May Allah Ta' Ala accept this from you!
Thank you! I’m glad it’s helpful
Thanks for covering this concept on your channel! Sylow’s Theorem is one of the few topics we did not cover in my Abstract Algebra class!
Most classes don't get to it. And many people who do learn it tend to have no idea what's going on cause it's usually super unclear!
Thanks for watching Eleazar!
It is the way to classify small finite groups. All the homework after is like "show what possible groups of order 75 up to isomorphism" and then your factor and apply the theorems.
Honestly, before you even started talking about the Theorem, I subscribed. Great video and and thanks for showing regular examples, I needed to understand this for my final.
I miss this stuff. I took Abstract Algebra 1987-1988 to prepare for graduate school and then again 1988 to 1989 in graduate school. So it's been 34 years. I definitely need a refresher.
That was very nicely explained.
Could you please explain from 5:38 to 6:34, why one of the orbits of G on Omega has size not a multiple of p? Thanks
Note that Omega is a disjoint union of its orbits, this implies that the size of Omega is the sum of the sizes of each orbit. Since Omega is finite, there are certainly a finite number of orbits and the size of each orbit is finite. If O_1,...,O_r are all the orbits of the action, then we can express that fact as |Omega|=|O_1|+...+|O_r|.
Suppose every orbit O has size a multiple of p. Then we can write for each i=1,...,r the following: |O_i|=p·h_i for some natural number h_i.
Thus, we have that |Omega|=ph_1+...+ph_r=p(h_1+...+h_r), meaning that |Omega| is a multiple of p.
But this is a contradiction because we have established by the combinatorial lemma that |Omega| is congruent to m modulo p, and by hypothesis, gcd(m,p)=1, so definitely not a multiple of p.
Wonderful! First time I getting it. I like the technique of keeping relevant information on the side of the board. I'm looking forward to The Two Towers and The Return of the King!
Thanks! Definitely enjoy 😃
Such a great well explained class! thanks!
Thanks! Definitely check out parts II and III and today's post.
Masterclass ^^
A bit confused on the part where you said that one orbit of the action of g on omega has size not equal to a multiple of p? why should that be true?
Thanks for the clear explanation.
And by the way,of all other proofs of Sylow, this one really stands out for we never need to appeal to the class equation!
I really like group actions!
@@ProfOmarMath Yes,they are awesome
Nice. Very clear.
Thanks!
Please can i know in wich level in university they study this theorem i realy need the answer !!!
400 lvl first semester
I’m preparing for the algebra qual and this was pretty helpful! A much simpler proof than the standard inductive proof starting with cauchys theorem as a base case (although that gives subgroups of size p^j for all j
I give a guided version of that on an exam actually.
Great video !!
But around
ua-cam.com/video/xTCxmr4ISU4/v-deo.html , how to proove the congruence to m ?
Thanks
Around 6:21, I think you really want to say "multiple of p" instead of factor. Nice video though!
Good catch!
Hey nice video. I believe there's an error in the thumbnail where G ≤ H rather than H ≤ G
Thanks! I'll fix it now
But I think it's cauchy 'a formula not sylow can you please double check
I think this is the first part of the Sylow Theorems but maybe there are other names elsewhere.
I didn't get it 😞
The theorem has a lot going on in it! Feel free to ask questions