Thanks! I have a number theory proof of Euler-Fermat's Theorem: ruclips.net/video/ijT3pmmal00/видео.html, and I plan to add a proof involving group theory soon.
Thank you very much for the video! I don't know if I can get an answer for my question, but here's my question: How can we say that the property that g itself is an element of coset of G (=g is an element of gH) implies that all union of gH is equal to the G? I understood that g is in fact an element of coset, and distinct cosets are disjoints, but I guess I just need one more line of explanation. Thank you
every g produces a corset of H. Not every corset is unique though so that when u have a total of n corsets some n-k are the same. So the disjunct union of the k unique corsets is the same as the union of the n corsets produced by G. since every corset contains g itself it is obvious that each g is contained in the union of all corsets.
in the sumation the indexes are up to k which is the number of cosets of h over g. before you demonstrated that two cosets have elements in common then they must be equal that is g1H=g2H. so what happens when for any reason there are repeated cosets. lagranges formula does not take into account that situation since the indices are up to k. I am a litle bit confused on that. cause as I see it the option that g1H=g2H for two diferent elements of g then this two sets are equal but their indexes are took into account eitherway
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Awesome! Got stuck on understanding Lagrange's Theorem in my class and this proof made it easy to understand.
This was perfect. Thanks. Can you sometime do like a video about Euler-Fermat's theorem proof and more Group theory.
Keep it up!
Thanks! I have a number theory proof of Euler-Fermat's Theorem: ruclips.net/video/ijT3pmmal00/видео.html,
and I plan to add a proof involving group theory soon.
This make so much sense and meaning and can relate what your saying and doing but which to learn more and see more video
Thank you very much for the video! I don't know if I can get an answer for my question, but here's my question:
How can we say that the property that g itself is an element of coset of G (=g is an element of gH) implies that all union of gH is equal to the G?
I understood that g is in fact an element of coset, and distinct cosets are disjoints, but I guess I just need one more line of explanation.
Thank you
every g produces a corset of H. Not every corset is unique though so that when u have a total of n corsets some n-k are the same. So the disjunct union of the k unique corsets is the same as the union of the n corsets produced by G. since every corset contains g itself it is obvious that each g is contained in the union of all corsets.
@@Brien831 What a great explanation! Thanks
in the sumation the indexes are up to k which is the number of cosets of h over g. before you demonstrated that two cosets have elements in common then they must be equal that is g1H=g2H.
so what happens when for any reason there are repeated cosets. lagranges formula does not take into account that situation since the indices are up to k.
I am a litle bit confused on that.
cause as I see it the option that g1H=g2H for two diferent elements of g then this two sets are equal but their indexes are took into account eitherway
Your videos are good for group theory
Please suggest me a book. For abstract algebra
Thank you for your help.
I teach from this nice open-source book: abstract.ups.edu/. If you master everything in this book then Dummit and Foote is a good choice.
When you said, in the first lemma, g1h=g2h' we can't conclude with this ? We don't have g1H include to g2H thanks to that ? Thank you for your help.
Absolutely brilliant
Maybe it should say a complete *and minimal* list of left cosets?
They are distincts, kind of redundant to say that