Abstract Algebra | Lagrange's Theorem

We prove some general results, culminating in a proof of Lagrange's Theorem.

[petargameplay]

This was perfect. Thanks. Can you sometime do like a video about Euler-Fermat's theorem proof and more Group theory.
Keep it up!

[abnereliberganzahernandez]

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

[liseg]

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.

[chasebrower]

Awesome! Got stuck on understanding Lagrange's Theorem in my class and this proof made it easy to understand.

[karankk]

Your videos are good for group theory
Please suggest me a book. For abstract algebra
Thank you for your help.

[georgettebeulah]

This make so much sense and meaning and can relate what your saying and doing but which to learn more and see more video

[하정훈-jj]

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

[FrankBria]

Maybe it should say a complete *and minimal* list of left cosets?