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
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 !
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
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
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
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Really wish these videos did not end 5 years ago. They are very good.
There are quite a few playlists on RUclips devoted to Modern Algebra but this is the best. Thank you so much for making and sharing this.
Wow what a explanation... you are also the one that make me love math
Great videos. I just wish you could produce new videos as fast as my Abstract Algebra course moves...
Thank you
These videos are so helpful.
Thank you very much.
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
Your explanation and your voice both are fabulous.
So clear and interesting. I just wish you could produce new videos Abstract Algebra.
I do plan on making more. I just wish I had some more free time!
Not sure if my professor would accept a verbal “ yes it is associative “.
it was clear , dats y attractive n not boring . well done .
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?
It is an really useful video.... thankyou sir 😊
Mr. learnifyable thank you very much , but how we can get this book or pdf
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 !
+Devesh Sawant Yes! It also has applications in physics (Lie groups) and chemistry (point groups).
very very inspiring. thank you so much Sir
At 7:54 what is it ? Four mod four is zero,I didn't get it.. could you please explain it more clearly 🌝
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
You make me understand this course
Great . Can you make videos about binary operations .please
Plz how to prove that S4/V4 is isomorphic to s3
Really good, ❤️🙏
Please prove associative property
THANKYOU SO MUCH!!
Thank you ❤️
Thank you so very much
Great! Keep Making Such Videos!
I definitely plan on making more. I'm glad to hear that you enjoy them!
Question:
How would I prove that for H1 != G and H2 != G, that H1 is not contained in H2?
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
Thank you so much, sir!
Very helpful......thanks
Thank you so much.
It is really helpful!
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
Yes, I certainly would like to make more videos. All of the topics that you mentioned are things I would like to cover.
"not just the identity"
So u mean, no trivial subgroups are including the identity?
Sorry I just confused... I hope u noticed
I mean non trivial proper subgroups
thank you this was very helpful
Awesome
this is one of those vids where you have to play it in x1.75 .....
You don't prove anything.
This is indeed associative without a prove.