(Abstract Algebra 1) Definition of a Subgroup

[@dr.dickie1418]

Really wish these videos did not end 5 years ago. They are very good.

[@jpdemont]

There are quite a few playlists on KZbin devoted to Modern Algebra but this is the best. Thank you so much for making and sharing this.

[@vigneshjothilingam7624]

Wow what a explanation... you are also the one that make me love math

[@Jkfgjfgjfkjg]

Great videos. I just wish you could produce new videos as fast as my Abstract Algebra course moves... Thank you

[@soklylorn7665]

These videos are so helpful. Thank you very much.

[@Mannu_81]

Your explanation and your voice both are fabulous.

[@samchau3476]

Please make abstract videos on Isomorphisms, Rings and Integral Domains. And more videos on how to get hold of one's intuition for making proofs. Most students suck at proofs and profs seem to think we should be able to do this shit or run after them all day

[@vismayanm2399]

It is an really useful video.... thankyou sir 😊

[@karthikaa2411]

So clear and interesting. I just wish you could produce new videos Abstract Algebra.

[@learnifyable]

I do plan on making more. I just wish I had some more free time!

[@michaelturkson3693]

very very inspiring. thank you so much Sir

[@jacklinesalome3753]

You make me understand this course

[@dipankardutta7823]

it was clear , dats y attractive n not boring . well done .

[@malcolmlamya8770]

Really good, ❤️🙏

[@juanjaimescontreras1798]

Not sure if my professor would accept a verbal “ yes it is associative “.

[@nugusujemal2317]

THANKYOU SO MUCH!!

[@11hitmanDagenius]

So clear and interesting ! thanks a lot for such videos ! Group theory has lot of applications right? I noticed it has lot of importance in other areas of mathematics . For example, there is an elegant proof of Fermat's Little theorem using abstract algebra !

[@learnifyable]

+Devesh Sawant Yes! It also has applications in physics (Lie groups) and chemistry (point groups).

[@valeriereid2337]

Thank you so very much

[@retamililu]

Mr. learnifyable thank you very much , but how we can get this book or pdf

[@milliekim5072]

Thank you so much, sir!

[@ardrajithendran9303]

Thank you ❤️

[@eslamababneh7094]

Great . Can you make videos about binary operations .please

[@iaktech]

Great! Keep Making Such Videos!

[@learnifyable]

I definitely plan on making more. I'm glad to hear that you enjoy them!

[@diribaaboma8341]

Thank you so much.

[@hamzarana7805]

Very helpful......thanks

[@ivankristoffergamilla5948]

In 8:25 why is it groups when you cannot have inverse of 2 it should have a value of -2 right to satisfy the inverse property?

[@xiaoboyali7906]

It is really helpful!

[@cavelinguam6444]

Awesome

[@bigboss2998]

thank you this was very helpful

[@tayyabazeb9270]

Plz how to prove that S4/V4 is isomorphic to s3

[@Love_Hope_from_Above]

Mr. Learnifyable: Another fine video on subgroups. I really like the examples and your teaching style is easy to understand. Will you be producing videos on cyclic groups, generators of a group/subgroup, cosets, permutation groups, etc. in the near future? You are a marvelous communicator! > Benny Lo California 2-17-2014

[@learnifyable]

Yes, I certainly would like to make more videos. All of the topics that you mentioned are things I would like to cover.

[@revanthreddy3790]

At 7:54 what is it ? Four mod four is zero,I didn't get it.. could you please explain it more clearly 🌝

[@davidnaray8398]

The answer to mod is the remainder So 5/4 is 1 and 1 remainder, so 5 mod4 is 1 Because 4/4 is 1, being evenly divided, it has remainder 0 As such, 4mod4 is 0

[@nossonweissman]

Question: How would I prove that for H1 != G and H2 != G, that H1 is not contained in H2?

[@raybroomall8383]

Without a definition of the operator != it is not clear what you mean, further if H1!=G then every element in G is in H! regardless of the operator !=. If H2!=G then every element in G is in H2!, therefore H1!=H2!. State that H1! = G and there n elements ksub 1 through ksub n-1 in G are in H1! therefore H1! is a proper subset of G. State that H2! = G and there n elements ksub 1 through ksub n-1 in G are in H2! so H2! is a proper subset of G state that e the identity element is in both H1! and H2!. So if there is an element in H2! that is not in H1! then there is some element r in H2! where e ! r is not in H1! k. But we said H1! =G then r is not in G and likewise cannot be in G because as we stated H2!=G So H1!=G=H2! ...qed

[@rajeswaripadari8543]

Please prove associative property

[@rikkiflows3867]

"not just the identity" So u mean, no trivial subgroups are including the identity? Sorry I just confused... I hope u noticed

[@rikkiflows3867]

I mean non trivial proper subgroups

[@mohammedalzaben7206]

this is one of those vids where you have to play it in x1.75 .....

[@kniinortey31]

You don't prove anything. This is indeed associative without a prove.