Lagrange's Theorem and Index of Subgroups | Abstract Algebra

[@moosophy1495]

You are going to save so many students i swear, thank you so much!

[@WrathofMath]

Thank you for watching! Really happy with how the Abstract Algebra videos have been coming out, looking forward to making it the definitive series on the subject!

[@punditgi]

Lagrange would be proud of this lesson!

[@WrathofMath]

Thanks Ezra, always want to do the big guys proud!

[@erinmeyers-t9w]

I remember reading this proof in my lecture notes, it was SO hard to follow... you've made it so simple!

[@chimetimepaprika]

Dude, this channel should be way more popular.

[@WrathofMath]

Thank you, I've worked very hard on making these Abstract Algebra lessons top notch, and look forward to continuing the playlist to make it the definitive series on the topic.

[@user-nb5ui1oq8j]

one of the best videos on lagranges theorem out there. thanks so much!

[@WrathofMath]

Thank you! I am trying to make this Abstract Algebra playlist the definitive series on the topic, and will be working on it a considerable amount this summer.

[@raiza.chakufora5271]

How do you find the index if given the cyclic subgroup with a permutation(123)?

[@MrCoreyTexas]

I was thinking some more about this and I don't know of any example where the identity is not 0 (for an addition style binary operator) or 1 (for a multiplication style binary operator). Theoretically you could have the identity element be 2 or 3. I'll probably write some programs to see what I can come up with.

[@aashsyed1277]

Can you give me the link to the video about order of a group ?

[@MrCoreyTexas]

Anybody have an example of a counterexample to the converse of lagrange's theorem? This one will be in the back of my mind for a while. I was thinking of the Klein group {1,3,5,7} with binary operator multiplication mod 8 and identity 1, I thought of expanding it to {1,3,5,7,11,13} with binary operator multiplication mod 14, but that's not a group because if you write out a Cayley table some of the elements don't have inverses (but the binary operator is closed). You appreciate how hard it is to come up with groups trying things like this. I think since 12 has a lot of divisors that would be the way to go looking for a counterexample, since you could look for a subgroup with 2,3,4, or 6 elements that doesn't exist.

[@Senorita-M]

Hello, thank you for this helpful video! Any advice to get full mark in this course?

[@fromtaoyuanhsinchyyou9075]

Thank you❤

[@WrathofMath]

My pleasure, lots more abstract algebra lessons on the way!

[@itachi6336]

wait so does it go both ways i.e. if G is a finite cyclic group then its order is a prime number and vice versa ?

[@WrathofMath]

Good question and no - it's only a description of groups of prime order, it is not a complete description of all finite cyclic groups. All the mod additive integer groups are cyclic, generated by 1, and one such as Z4 for example has composite order.

[@mihaelashumarska3698]

amazing

[@WrathofMath]

Thank you!

[@dr.anubhavkumarprasad5765]

How to proof set with single element is a group ?

[@erinmeyers-t9w]

The only group with only one element is the trivial group (only contains the identity)