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Cosets in Group Theory | Abstract Algebra
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- เผยแพร่เมื่อ 3 ธ.ค. 2022
- We introduce cosets of subgroups in groups, these are wonderful little discrete math structures, and we'll see coset examples and several coset theorems in this video. If H is a subgroup of a group G, and a is an element of G, then Ha is the set of all products ha where a is fixed and h ranges over H. Ha is called a left coset of H in G, and the right coset aH is defined similarly. We will see finite cosets and infinite cosets. #abstractalgebra #grouptheory
Cosets of a subgroup partition the whole group, and cosets with common elements are equal, we discuss both of these facts in this video, and we prove the latter fact. Our discussions of cosets will eventually lead to a proof of a famous result: Lagrange's Theorem.
Cosets of a Subgroup have the Same Size: (coming soon)
Proof that Cosets Partition the Group: • Proof: Cosets Partitio...
Lagrange's Theorem: (coming soon)
Abstract Algebra Course: • Abstract Algebra
Abstract Algebra Exercises:
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This was clear, easy to understand and concise without leaving things out. It was perfect. Thank you!
Thanks a lot for watching and the feedback! These concept overview lesson are the hardest to make, as it takes a lot of time to decide exactly what to cover in them. I just finished recording the follow up lesson proving that cosets partition groups, a great result!
Intro to Abstract was probably my favourite class in my undergrad studies.
Same!
Very coherent. This is a great format for going over definitions. Since you clearly scripted it as well the video was pleasant to listen to (a lot of similar videos tend to "improvise" on the fly, but that doesn't work as neatly from my experience and is fairly error prone).
Excellent video!
Thanks so much! As I work on this series, I have tried to make these the highest quality abstract algebra videos available. I cannot match the production value of Socratica, but I hope I can make up for that in thoroughness. And especially for overview type videos such as this one, extensive preparation is required to make the lesson smooth and comprehensive. Thanks for watching!
Glad to have you back in Richard Rusczyk Hoodie form!
I liked your morning coffee presentation style as well.
The result from abstract algebra that is most memorable to me is the shockingly easy proof that every group is isomorphic to a permutation group(operation composition)
Amazing video. I was being thought about this using equivalence relations but this is far far more intuitive and now I actually understand what’s going on instead of just regurgitating formulas!
Greatly appreciate your work and quite comprehensive explanations. Very brief and straightforward thank you!
So glad you found it helpful! Thanks for watching and let me know if you have any questions!
Thanks for this video, I was watching another series and got stuck on cosets. The examples helped big time!
Awesome, thanks a lot for watching!
Great introduction to group theory. Please keep on good work !
thank you so much for everything that you do!
I've been watching your videos since the beginning of my studies with calculus 1, and I always find myself so relieved when a course I'm doing has a relevant playlist/video in your channel (:
your channel is a place to find sense and great explanations for me when everything is spiraling. so thanks a lot, and hopefully for many more encounters here in my next courses! (:
Thanks so much for the kind words Dana - I'm so glad my videos have been helpful for you and I hope they continue to be! Let me know if you ever have any questions!
Thank you for these videos.
Very clear.....nice!
Not only is this video really great , it is coset great .
Thank you!
Thank you. The motivation was enigmatic and confusing to me in the beginning.
Glad to help, cosets are fascinating!
Thanks for the short but sharp explanation. Was struggling to understand cosets while reading the textbooks.
Glad to help! Thanks for watching and check out my Abstract Algebra playlist if you're looking for more! th-cam.com/play/PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN.html
wow wow wow!!! great job sir. very easy and simple to understand.
Thank you, glad it helped!
Bless your heart, I finally understood, just in time for exams. I hope I make it this time 😤
Thanks for watching and good luck!
Thanks man ...Great explanation👍
Glad it helped!
Thank you so much! Greetings from Germany
You're very welcome! Salutations back from Cape Cod, USA!
Thanks for your help man
Been struggling with it ❤
Glad I could help! Let me know if you have any questions, I'm continuing to work on new Abstract Algebra lessons.
@@WrathofMath Alright bro, appreciate it
I subscribed to your channel 😊
Can you please prove that Ha = {x€G: x is congruent to b mod H}
your eyes are too good and your explanation too. And hello, I am from India.
Thank you very much! Glad it helped!
this guy doesn't blink
Is this definition available in code theory?
Hey love your videos! Is there a way to get all of what you wrote in these videos?
Thank you and good question! I had never considered making my actual writing from the videos available as a PDF or anything until the video I did recently solving all the 2023 AP Calc free response questions. I made that available for free on my Patreon page; and I might consider making all notes available for Patreon members - it takes some extra work to get it all cleaned up and ready to share as a PDF otherwise I'd have probably started doing it for all videos by now. If there is a particular video you'd like the notes from let me know and I can reply with a link, but it will be a project to start sharing them all.
nice hoodie
very nice
Thsnk you!
thanks
My pleasure, thanks for watching!
In this example 2:50, shouldn't you write the classes of the elements? i.e. Z4 = { [0], [1], ... }
This could all be derived by the fact that the coset is a equivalence relation, right?
I'm not sure what you mean by "this all", since I don't remember everything I included - but probably!
This was very clear and well done, but I'm still a little confused about the end and hoping someone can help me. You used the fact that Hb is a subgroup to apply associativety, but then stated that only one coset can be an actual group (because it's the only one with an identity). So my question is, how are you allowed to assume that Hb has associativety if it's arbitrary and may in fact be a coset that is not a subgroup?
Thanks for watching and good question! I didn't use the fact that Hb is a subgroup, since it may not be. I used the fact that the group operation is associative on all of its elements. We may narrow our focus to a subset of those elements which does not form a group (such as a coset) but the operation is still associative on all of those elements. Does that make sense?
@Wrath of Math Yes, it makes sense now. The cosets are subsets of the entire group, and properties like closure and associativety permeate throughout the entire group so they affect the subsets but the identity isn't an operation so it doesn't "move around". At least, that's what I'm taking away from it.
Right, the operation has certain properties with regards to the groups elements, and those properties remain whether we look at all the group elements,.or only some of them.
@Wrath of Math Awesome. One last question, and excuse me if this comes off nonsensical as I'm only just learning this stuff, but could I test for isomorphism of two groups by replacing one groups "identity coset" with the others and vice versa? I imagine if closure and associativety is assumed then showing the identities are interchangeable gives way to isomorphic groups?
I'm not positive I understand your question, but if I understand correctly the answer is no. For example, the reals are not isomorphic to the complex numbers under multiplication, yet they both have the rationals as an "identity coset". Thus this is an example of two groups with isomorphic identity cosets (even more, they are exactly equal cosets) yet the groups are not isomorphic.
G ={0, 1,2,3, +} is not a group because it is not closed as 2+3 = 5 but 5 is not an element of Z. Please explain.
The operation is not plain addition, it is addition modulo 4. So 2 + 3 is equal to its remainder when divided by 4, which is 1 since 5/4 has a remainder of 1. Similarly, 3+3 is 2 mod 4 because 3+3=6 divided by 4 has a remainder of 2.
13:36 don't say 0 say e (general identity
We all know your elocution and delivery would be perfect, so rather than mouthing off here, see if you can do a better job on your channel.