Normal Subgroups and Quotient Groups (aka Factor Groups) - Abstract Algebra
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- Опубліковано 14 гру 2018
- Normal subgroups are a powerful tool for creating factor groups (also called quotient groups). In this video we introduce the concept of a coset, talk about which subgroups are “normal” subgroups, and show when the collection of cosets can be treated as a group of their own. As a motivation, we will begin by discussing congruences.
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Dummit & Foote, Abstract Algebra 3rd Edition
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The most useful series of mathematics videos I have encountered since 3blue1 brown
Yup. If you know any other awesome series like this, then it'll help me a lot.
I agree
Check out Faculty of Khan as well.
@@randomdude9135 please check Mathdoctorbob's series on abstract algebra... It's really great, really intuitive, and goes into phenomenal depth.
@@mychannelofawesome thanks!
Everything they do here concerning teaching is badass., meaning they look "bad" in front of most professors because their biggest fear is - paradoxically- to be understood while the greatest mission of Socratica is to appear understandable. And hardly a one does a better job. Because maths is first and foremost supposed to be one thing: intuitive and fun. and ONLY THEN to be formal but only AFTER one has established and examined the concrete cases. Maths then appears to be a collection and characterization of those examples and not a collection of dead and unmotivated formal arguments, definitions and theorems.
Formal symbols do have phenomenal value but only if one has gotten and intuitive understanding of the theorems and definitions first. Socratica does exactly this. That's why she should be nominated the Oscar prize for teaching mathematics.
“A living Socrates”
Winston Jiang yeah,she kinda is :)
I blame the subject more than the teaching. It's very difficult for me to relate abstract algebra to anything I've seen in the past. The only interest I have for it now is it appears to be central in understanding why there is no general solution in radicals to the quintic equation. Which is interesting, but man, do we have to learn ALL this stuff just to understand that one problem?
@@theboombody
Linear algebra is abstract algebra, as a vector space is an abelian group with a compatible field action (scalar multiplication). So in a sense, anything you can use linear algebra for is an application of abstract algebra. That aside, the slight generalization of vector spaces (where the field may be weakened to a ring), called modules, appear in calculus on manifolds: the set of vector fields on a (real) manifold M forms a C^r(M,R)-module.
A formal theory of polynomials and rational functions also falls under abstract algebra, in the form of rings and fields. Polynomials are more than just "which ones can be solved via explicit formula" though; for example, differential equations such as y'' + 2y' + y = 0 can be studied as polynomial differential operators e.g. D^2 + 2D + 1. This is, of course, trivial for the constant coefficient case, but when the coefficients are polynomials, you end up with a polynomial ring which is not commutative, and so different techniques need to be developed.
Groups themselves also find a good amount of importance in calculus and differential equations on manifolds, in the form of Lie groups. Lie groups are groups, first and foremost, which also have some kind of (smooth) manifold structure. Their related objects, the Lie algebras, are vector spaces with a certain kind of vector product (for example, R^3 with cross product is a Lie algebra). It is precisely the properties of groups that make Lie groups so useful, either as a manifold of study or as the typical fiber in a principal bundle structure.
One last thing, the quotients that are being developed in this very video are the basis for the major tensor algebras, including the exterior (Grassman) algebra and the symmetric algebra. The tensor product of vector spaces itself is constructed by taking the vector space whose basis is indexed by pairs of vectors, then taking the quotient by the ideal generated by the properties we wish to hold. From the complete tensor algebra, the exterior and symmetric algebras are achieved by taking the quotient by the ideal generated by skew-symmetric and symmetric multiplication, respectively. Ideals are simply the equivalent of normal subgroups for rings and similar contexts, basically those substructures which allow quotients to have the desired structure.
This is just a small sample of the use of abstract algebra in other areas of mathematics, obviously localized to my particular area of study. I hope you can come to realize that abstract algebra is not as self-contained as it seems, and the techniques and language learned from studying the subject is of great importance even in the relatively grounded subjects of calculus and differential equations.
@@alxjones what is your area of study/research?
I had to play the video multiple times with several pauses along the way for me to grasp the concept.
my head hurts :( but i will try to watch it again later :)
Me too
lmao same
This time I understood atleast 50% I think. Time to ponder on my own and scribble around in a book.
SAME! I'll come back to the ending later....
I am already quite old and trying to learn abstract algebra. Sometimes I just need a very clear and down to earth description of a mathematical object which can be quite hard to teach yourself from a book This channel provides an excellent tool in that regard. Thank you!
Now try Analysis or Calculus III and absolutely tear those last remaining hairs on your head out. I took a second master in Electrical engineering when I was 32 and I felt like a fucking grandpa already.
I find it helps to have different source material for the same subject, and to skip back and forth between sources. These videos are great for that purpose.
This playlist is awesome!
TOPOLOGY WHEN?!?!? 😁👍🏻
S... waiting
Can watch this almost effortlessly in the evening, trying to read the same theory from a book took almost a week of studying every morning and led to a more superficial understanding than this video. You guys are geniuses when it comes to presenting ideas, you're definitely on the list of channels I'd like to support when I'll be able to.
In Motivating Example, How do we get remainder 1 if we divide -14, -9, -4 etc. by 5? Please reply i am so confused 😢 integer mod 5 is confusing me
@@homiramanuj The smallest number closest to -4 that is divisible by 5 is -5 so -4 - (-5) is 1. Same goes for -9, -14 and so on.
@@homiramanuj In the division quotient can be negative numbers. Thus, by dividing -14 by 5 we get -3 as quotient and -14-(-15)= 1. The point here that the quotient times divisor should be less than or equal to dividend.
This was my first time trying to learn and it didn't help. But I'm gonna try again, and again, and again until it makes sense. I'm committed to finishing your playlist with usable understanding. Keep up the amazing work, Socratica team!
How did it go ??? Am starting on a similar journey
Somehow I stumbled upon your channel while searching for the Primer Vector Theory a couple days ago, and then watched your entire astronomy series ... and here I am watching the entire Abstract Algebra series.
One thing that has always frustrated me trying to learn these things from, say, Wikipedia is that they're always written by people who fully understand the subject FOR people who fully understand the subject and and are quite difficult to understand until you understand it - even in cases where the concepts are quite simple.
I'm so glad to have found your channel where you explain things so simply and so clearly. Thank you so much!
From 6:00 onwards, although the real case is more general, the entire thing becomes a lot easier to understand if every time she says "times" or "multiply" you think "plus" or "add", every time she says "N" you replace it with "Modulo(someNumber)", "e" is "0Modulo(someNumber)", and "x" and "y" are just numbers that aren't a multiple of someNumber. Cheers!
What an amazing series, this is a goldmine! Perfect depth and speed.
One of the best math series on youtube. maybe the best if you already have enough background to understand this. Thank you for doing this !
The way she teaches and explains , totally incredible...!
The way you simplify a complex concept is great!
You know it's a good video when the content seems simple and is really easy to comprehend. Sometimes I lose myself in all of the new definitions etc. in my Algebra course, but these videos pull everything together and help greatly with the motivation behind everything you learn.
Excellent presentation: clear, to-the-point, fluid.
the most approachable abstract algebra course online. thank you so much for your hard work!
I use this series to accompany my lectures on abstract algebra, it helps me so much to understand what is going on. Thankyou!
Was self studying Galois Theory and this helped to recap a lot of forgotten theorems, thanks a lot!
Trying to find a word that describes my gratefulness for such incredible explanative videos.
Group is [ i (identity) , r1 (rotation 1/3), r2 (rotation 2/3), s1 (sym 1), s2 (sym 2) , s3 (sym3) ]
(i,r1,r2) is a subgroup.
This subgroup is normal because:
s1* r1 *s1 =r2
s2* r1 *s2 =r2
s3* r1 *s3 =r2
(a symetry is its own inverse element)
I got this too!
In Motivating Example, How do we get remainder 1 if we divide -14, -9, -4 etc. by 5? Please reply i am so confused 😢 integer mod 5 is confusing me
eh meshe
In really like the presentation style. Everything is very clear and all the explanations are easy to follow. Thank you so much
@11:04 the set of permutations (123) (132) and the identity permutation form a normal subgroup of S3
isnt it the only (non trivial) sub group as well as the only normal sub-group? any other basically fail to endure the closure property
@@Yougottacryforthis it’s the only non trivial normal subgroup, but not the only non trivial subgroup, just pick a set containing the identity and a 2 cycle
@7:25 replace y with y^(-1)
This series is blowing my mind, your work is highly appreciated
In Motivating Example, How do we get remainder 1 if we divide -14, -9, -4 etc. by 5? Please reply i am so confused 😢 integer mod 5 is confusing me
The best video i ever viewed on youtube about group theory. Thanks alot
Too good explanation which covers most important part of normal subgroup. U are truly a good teacher.
I really like you because explain the subject in an easy and understandable way.
Mam your way of delivering lecturer is amazing,outstanding.. God bless you
This was explained very amazingly. Thanks for this :)
Damn I really needed this video 4 days ago before my exam!
(it went ok but factor groups was something that went over my head for most of the semester)
One more bit of constructive feedback, the exercise at the end, "find a normal subgroup of S_3", assumes knowledge of what symmetric subgroup S_3 is --going by the Abstract Algebra playlist order, the concept of a symmetric subgroup hasn't been introduced yet.
The most understandable videos of abstract algebra on UA-cam.Very easy to understand
In Motivating Example, How do we get remainder 1 if we divide -14, -9, -4 etc. by 5? Please reply i am so confused 😢 integer mod 5 is confusing me
@@homiramanujbecause -4 = (-1)*5+1
Thank you. Excellent presentation.
These videos are so helpful it's unreal
your all videos are very descriptive .it helps to solve many troubles .i wish to do more and more videos. thank you
So so great. So well delivered.
Thanks! Had to watch in 0.5x the speed to hang on, but very helpful! :)
WOW, so nice and easier to understand. Beats the textbook 100%.
Great explaination love it , makes the topic fun 💝💝
Excellent work. Students are recommended to watch this video. It will help to motivate you properly.
Thanks a lot this videos series is very useful. It explains everything in a very simple way🙂🙏🏻🙏🏻.
Outstanding video lecture. This video lecture is very helpful for self-study.
Thank you so much for the explanation!
she saved my whole fxxking life during the senior this fall
Very good work : still to give a constructive critic: 1) I think the argument for definition yN=Ny could be exposed without going down to elements and avoiding inverse as much as possible.2) The transition from Z,+ to multiplicative is not the best though I cannot think of a simple multiplicative example fro cosets.3) It is so nice to finally see questions being asked to motivate a definition. Still from a didactic point of view it could be worth repeating the question at end ( recap style).
Thanks a ton !!! Explained with such clarity. It was to the point , excellent explanation.❤️
Is it nityananda who's in your profile?? 🤔
@@sudarshann7194 eh ktir excellent
When we will have topology series like abstract algebra ?
i wish they do it
Would you want me to ? Or in other words: would it still be useful for you ?
@@howmathematicianscreatemat9226 yes!!
@@_qpdbdbqp_ okay, till when do you still need it? Tell me the date and also if you think good explanations could help your classmates too? If you tell me, then maybe I'm gonna start producing them when I'm back from holiday on the 25th of February. You would then view your first plesant set-topology video on the 27th of February. But if you want to me to start, confirm your request.
@@howmathematicianscreatemat9226 I'd also be grateful if you began posting on 27th Feb
*Love the way you teach*
So well explained!!!! Thank you! I have an exam tomorrow. I have now more confidence than apprehension XD
wow you took such a complicated subject and make it so simple.
studying for my math classes is enjoyable with Socratica
THANK YOU SO MUCH FOR THIS
This is so helpful, I can now clearly visualise these concepts 😍🙌. Your videos are amazing, Thankyou Socratica✨
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Very helpful videos. I had to pause a lot to understand but worth it.
that was..... amazing. great job
Very good. Thank you so much.
took me watching it twice to understand perfectly(have to oil my brain)....awesome to the point explanation.
This is our favourite thing about UA-cam compared to classes - you can just rewatch! Thanks for sticking it out with us! 💜🦉
I love the way you teach
Thank you so much!!!
Such a beautiful topic
You people are amazing!
thank u i love the way u teach. I didnt understand my professor but here go everything I needed any videos on Mathematical Analysis?
Happiness is nothing but understanding stg properly. These vids series are fuckin' awesome.
Where were you in 2016 when I was taking Abstract Algebra??? 😝
Love the series. I’ll be going through each one several times until I understand your proofs and can duplicate them (again?).
great and very well done. Congratulation.
bro this video is amazing. i was like wtf is this quotient groups and cosets reading dummit&foote. came here and its all clear now. can continue reading. thanks
Need to watch More videos
😃
Thanks. Great video. !!
great video, thanks a lot!
Very interesting explanation...!!!!
You are my best teacher.
So how would one prove the second part of the statement at 7:27? I proved it by showing the first part, and then showing that the two have to have the same order and no duplicates. I'm not sure if this is the right approach though.
it's very helpful to me .
thank you .
Pls make more videos on abstract algebra i love your explanation very much
Simple groups are the primes of group theory.
S(3) is isomorphic to D(3) the dihedral group of 6 elements so the normal subgroup would be the rotations or the subgroup generated by a 3-cycle.
Yes, I agree, and it is easy if you see that all inverses out of rotation subgroup are itself. f1*g1*f1=g1
these videos are awesome!
saving my time & leisure time
I wish I could upvote this video 100 times.
Wow, great video! I learned a lot. One thing that felt unexplained was this statement just before 10:00 about factor groups that "the inverse of x⋅N is x^(-1)⋅N". I can play around with the integers mod 5 as an example and see it's true, but I'm wondering how to convince myself it works in general. Thanks again for making these :)
N is invariant subgroup, it means that for any x xN=Nx. (xN)(x^(-1)N)=(Nx)(x^(-1)N)=N(xx^(-1))N=N1N=NN=N. In factor group N is 1.
My favourite teacher
Yes! More Socratica.
one can listen this forever/..
@11:03 the set {I, (123), (132)} is a subgroup of S3
@10:45 if you find the factors associated with the composition series: is it possible to then reconstruct the group after factoring it
Precisely in what sense is the "product" of G/N and N "equal to" G? (Which is sort of what one would expect from a quotient, by analogy with ordinary numerical fractions.)
bye socratica, i watched your video on normal subgroups& quotient groups. very beautiful presentation & interesting maths subject. it is my request tu pl relaese vedio on "boolean algebra & it's applications" thank u
I think reasoning on 6:12 lack some crucial point:
You can use any element from gN coset to generate gN coset, i.e. if h in gN coset then gN=hN. (and looks like it is not a property but rather a definition of these equivalence classes(cosets): "the set G/N is defined as the set of equivalence classes where two elements g,h are considered equivalent if the cosets gN and hN are the same" brilliant.org/wiki/quotient-group/)
That's why we can use xyN coset on the right instead of generic zN coset (since x in xN and y in yN: zN should contains xy element to respect definition given at 4:19, and then xy can be used to generate zN coset which means zN=(xy)N)
That's great! I bet you can do one on type theory.
Thanks again
Salute to you guys 👌
It gets confusing when she says "For cosets to act like a group xN yN = xy N"
I didn t understand why so I thought about it.
Assume the opposite: xN yN =zN with z not equal to xy.
Using the identity we get:
xy = zn for some n in N, multiply the left side by z^-1 we get
z^-1xy = n.
Therefor the coset x^-1zN has the element: x^-1z n = (x^-1z )(z^-1x)y = y
So the coset yN and x^-1zN have the element y in common which according to the socratica video about lagrange theorem is a contradiction because two cosets can t have elements in common.
This is why z must be equal to xy for the cosets to behave like a group.
Thank you for the explanation, although i didn't understand it very much. I think you could have said:
xy = zn for some n then y = (x^-1)zn
So y belongs to the coset (x^-1)zN, but y belongs also to the coset yN but the cosets have no element in common so itust be xy = z
@@ilguerrierodragone129 yes thank you.
Sometimes I have difficulty explaining
Or thinking straight haha
just in time for finals ;)
Hooray! That's what we were hoping. :D Good luck!!
@@Socratica i have a question:
since y^-1(N)y = N, if we multiply both sides by y in their left.
y[y^-1(N)y] = yN
Ny = yN
so does this mean that cosets form a group only if left cosets is the same as their right cosets? is this always the case?
@@CreolLanguag good question.
What's the answer of this question? Did you get it?
@@CreolLanguag Yes that is correct!
Watching this during finals
Very nice
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Do you plan to teach vector spaces in future? Your videos are incredibly helpful btw :) Thanks a lot
They have a video on it from three years ago.
you are amazing!!!