Normal Subgroups and Quotient Groups (aka Factor Groups) - Abstract Algebra

[@Socratica]

If you'd like to learn more, we have a free course on Group Theory! www.socratica.com/courses/group-theory

[@siddharthprakash8942]

The most useful series of mathematics videos I have encountered since 3blue1 brown

[@randomdude9135]

Yup. If you know any other awesome series like this, then it'll help me a lot.

[@manuthebroker5598]

I agree

[@vibodhj349]

Check out Faculty of Khan as well.

[@mychannelofawesome]

@@randomdude9135 please check Mathdoctorbob's series on abstract algebra... It's really great, really intuitive, and goes into phenomenal depth.

[@effy1219]

@@mychannelofawesome thanks!

[@howmathematicianscreatemat9226]

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.

[@winstonjiang3621]

“A living Socrates”

[@howmathematicianscreatemat9226]

Winston Jiang yeah,she kinda is :)

[@theboombody]

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?

[@alxjones]

@@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.

[@itszeen7855]

@@alxjones what is your area of study/research?

[@ChristopherKemsley]

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!

[@joshuaronisjr]

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!

[@ericvosselmans5657]

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!

[@ThefamousMrcroissant]

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.

[@samiaario8291]

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.

[@yeast4529]

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.

[@Arv.-]

The way you simplify a complex concept is great!

[@alexamadori9884]

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.

[@homiramanuj]

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

[@godspower_eze]

@@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.

[@erfanmirzaei705]

@@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.

[@cameronspalding9792]

@11:04 the set of permutations (123) (132) and the identity permutation form a normal subgroup of S3

[@Yougottacryforthis]

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

[@cameronspalding9792]

@@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

[@cameronspalding9792]

@7:25 replace y with y^(-1)

[@souravacharya]

The way she teaches and explains , totally incredible...!

[@IBMboy]

my head hurts :( but i will try to watch it again later :)

[@randomdude9135]

Me too

[@ScilexGuitar]

lmao same

[@randomdude9135]

This time I understood atleast 50% I think. Time to ponder on my own and scribble around in a book.

[@adeelali8417]

SAME! I'll come back to the ending later....

[@Gaspard832011]

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)

[@sammie1824]

I got this too!

[@homiramanuj]

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

[@Adam-r2n]

eh meshe

[@JPL454]

@@homiramanuj You always go for a number

[@ozzyfromspace]

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!

[@luyombojonathan7715]

How did it go ??? Am starting on a similar journey

[@alayamaryim]

The most understandable videos of abstract algebra on KZbin.Very easy to understand

[@homiramanuj]

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

[@ClumpypooCP]

@@homiramanujbecause -4 = (-1)*5+1

[@o.s.h.4613]

@@homiramanujA year late! But 5x1÷(-4) = remainder of 1. 5x2÷(-9) or 10÷(-9) = remainder of 1; 5x3÷(-14) or 15÷(-14) = remainder of 1. Etc. The sign doesn’t matter here, just that you need to make one step (in either direction) to get to the denominator :)

[@jonathanpopham5483]

the most approachable abstract algebra course online. thank you so much for your hard work!

[@sadied0g]

This playlist is awesome! TOPOLOGY WHEN?!?!? 😁👍🏻

[@bakkamydestination]

S... waiting

[@vyrsh0]

Have they uploaded it now?

[@Anna-jy7cj]

This series is blowing my mind, your work is highly appreciated

[@homiramanuj]

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

[@darsh9156]

This really helped me understanding this topic. I was really confused and now all the confusion is gone. Thanks a lot .

[@tomjoyce9401]

Excellent presentation: clear, to-the-point, fluid.

[@raunitsingh676]

Trying to find a word that describes my gratefulness for such incredible explanative videos.

[@ThePimp4dawin]

What an amazing series, this is a goldmine! Perfect depth and speed.

[@GGC728]

The best video i ever viewed on youtube about group theory. Thanks alot

[@merlijn1e]

I use this series to accompany my lectures on abstract algebra, it helps me so much to understand what is going on. Thankyou!

[@jeromejean-charles6163]

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).

[@林宇博-w9l]

she saved my whole fxxking life during the senior this fall

[@fitzregelbrugge3673]

I am studying for my Algebra exam and these videos give an amazing extra insight and perspective on the matter. Thank you!

[@Socratica]

We're so glad you're finding our videos helpful! Good luck with your exam!! 💜🦉

[@youtwothirtyfive]

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.

[@saeedahmedhashmi9448]

Mam your way of delivering lecturer is amazing,outstanding.. God bless you

[@carlsagan9808]

These videos are so helpful it's unreal

[@turokg1578]

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

[@cameronspalding9792]

@11:03 the set {I, (123), (132)} is a subgroup of S3

[@Bloodsaberxy]

Was self studying Galois Theory and this helped to recap a lot of forgotten theorems, thanks a lot!

[@shacharh5470]

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.

[@Nand0san35]

Yes, I agree, and it is easy if you see that all inverses out of rotation subgroup are itself. f1*g1*f1=g1

[@LOL091027]

When we will have topology series like abstract algebra ?

[@DiegoGonzalez-xl9us]

i wish they do it

[@howmathematicianscreatemat9226]

Would you want me to ? Or in other words: would it still be useful for you ?

[@_qpdbdbqp_]

@@howmathematicianscreatemat9226 yes!!

[@howmathematicianscreatemat9226]

@@_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.

[@John-js2uj]

@@howmathematicianscreatemat9226 I'd also be grateful if you began posting on 27th Feb

[@malenaalmasi1774]

In really like the presentation style. Everything is very clear and all the explanations are easy to follow. Thank you so much

[@youtwothirtyfive]

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 :)

[@vladislavnikolaev800]

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.

[@anotheryoutuberperson38]

You helped me so much! Thank you for these videos because my professor is not good.

[@rosaaanaaa]

studying for my math classes is enjoyable with Socratica

[@ramumaha2779]

took me watching it twice to understand perfectly(have to oil my brain)....awesome to the point explanation.

[@Socratica]

This is our favourite thing about KZbin compared to classes - you can just rewatch! Thanks for sticking it out with us! 💜🦉

[@eminentlostcity1754]

I'm a university student from India and here teachers don't teach basics because most of the students here are really good at studies....so this is really helpful for weak students like me.... I'm trying my best

[@ajsdoa6282]

Thanks! Had to watch in 0.5x the speed to hang on, but very helpful! :)

[@sujoydey8359]

Too good explanation which covers most important part of normal subgroup. U are truly a good teacher.

[@finn5571]

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.

[@sutanuhait3283]

Excellent work. Students are recommended to watch this video. It will help to motivate you properly.

[@RobotProctor]

I think N={123,231,312} is a normal subgroup (rotations of the dihedral group). 123 is e. The 3 possible elements to form cosets with are the flips from the dihedral group, which are their own inverses. So y*N*y^-1 will involve a flip, rotation, and a flip again which will result in a simple rotation. Since this is a rotation it's in N.

[@iwantaoctosteponmyneckbut3545]

Not really, but you have the right idea by focusing on 3-cycles (i.e. permutations whose cycle decompositions have 3 numbers in them). If N is a subgroup of S3, we know it uses the group operation of S3, function composition, which I'm writing as "*". (1 2 3) can't be the identity for N because (1 2 3)*(2 3 1) = (1 3 2) =/= (2 3 1). N needs the identity element from S3, which is just (). Also, the (2 3 1) and (3 1 2) in your N are the same permutation In reality, N = {(), (1 2 3), (1 3 2)}. You have () as your identity, and the other two permutations act as inverses of each other. Associativity and closure are pretty obvious too, so this N is a group, and thus a subgroup of S3

[@huttarl]

@@iwantaoctosteponmyneckbut3545 Where did you get (1 2 3)*(2 3 1) = (1 3 2)? By my figuring, John C is right that (1 2 3)*(2 3 1) = (2 3 1). (1 2 3) is indeed the identity element. Maybe you're using cycle notation? whereas John is using one-line notation.

[@user_sense]

@@iwantaoctosteponmyneckbut3545 you just defined the same thing as jonny but with cycles, and unless i missed something, none of you gave a proper proof, (but jonny did explain why it is indeed a normal subgroup)

[@tekhiun]

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 !

[@knok16]

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)

[@RobotProctor]

I wish I could upvote this video 100 times.

[@sarthakkrishna1737]

So well explained!!!! Thank you! I have an exam tomorrow. I have now more confidence than apprehension XD

[@exx8eran]

wow you took such a complicated subject and make it so simple.

[@ashmytom8927]

You have made things simple!....Thankyou

[@briancannard7335]

It's great how you show the truth of mathematicians. They love to multiply elements and cosets in the same statement jumping between levels of abstraction.

[@briancannard7335]

Why care that cosets form a group?

[@briancannard7335]

Mathematicians need to invent a disclaimer about generalization levels, and attach it to all concepts created. Each time I see a "Factor Group" now or a "Quotient Group" I'll think about, ah... it's a group of cosets, not of elements! Thank you @Socratica!

[@halilibrahimcetin9448]

Happiness is nothing but understanding stg properly. These vids series are fuckin' awesome.

[@syamalchattopadhyay2893]

Outstanding video lecture. This video lecture is very helpful for self-study.

[@Souley239]

I really like you because explain the subject in an easy and understandable way.

[@ThePharphis]

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)

[@tomwellington4255]

5 years have passed, has this knowledge been useful to you outside the classroom?

[@ThePharphis]

@@tomwellington4255 No, but I teach high school and extremely rarely do I ever touch on these topics while tutoring. I think only once or twice (and it wasn't this far in depth)

[@tomwellington4255]

@@ThePharphis Thanks for taking the time to respond!

[@raymondblake5765]

At 6:22 the video says that for the cosets to act like a group x*y should be in the product of xN *yN. What property of a group makes this so and why? There must be something I'm missing...

[@mike11022]

so that the product of two cosets is well defined. it's not a property of group, it's a necessary condition for the product of cosets to be well defined.

[@Socratica]

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[@ijyoyo]

WOW, so nice and easier to understand. Beats the textbook 100%.

[@ronycb7168]

Indian here really enjoying this series happy republic day guys

[@homiramanuj]

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

[@ueiwqoak]

I get lost at 6:05 and can't climb out of my confusion. I can't understand where x*y is an element of the multiplication of the cosets. What is the multiplication of the cosets?

[@Socratica]

If A and B are two cosets, then A*B is the set of all products a*b where 'a' is in A & 'b' is in B. With normal subgroups, something wonderful happens. The collection of cosets form another group!

[@randomdude9135]

@@Socratica Thanks for clarifying. It's kinda similar to the Cartesian product of two sets.

[@b43xoit]

@@Socratica That's kind of what I thought you might mean by the product of two sets in this context. So if a person were computing the product of sets, the Cartesian product would have |N|^2 elements, and all but |N|, you would have to throw out as leading to duplicate products. Next point of confusion (after not being sure what set multiplication was): At 6:09, you say that for cosets to act like a group, xy must be an element of (xN)(yN). How can it not be? We know that x is in xN and y is in yN as you had just demonstrated. So (x, y) must be in the Cartesian product of xN with yN. So xy must be in the set product.

[@reidchave7192]

@@Socratica How can the cosets form a group if they don't contain the identity element?

[@marcevanstein]

@@reidchave7192 My understanding is that there's an extra level of abstraction going on: we're treating each of the subsets of the original group (cosets) as *elements*, and it is these that form a group when you have a normal subgroup. So it's not a question of containing the identity element of the original group; if we had a normal subgroup, then it (the whole normal subgroup) acts as an identity element within this group of cosets, since when multiplied by any other coset it leaves it unchanged. I think.

[@youtuber_nr3504]

Why do x^{-1} and y^{-1} exist at 6:55? Cosets do not have inverse elements in general, do they?

[@sinisternightcore3489]

Hmm... Both x and y are contained in G, so they must have inverses.

[@192ali1]

Thank you. Excellent presentation.

[@estebanvasquez-giraldo5770]

Why is it that in 6:05, for the cosets to act like a group, x*y must be in (xN)(yN). I appreciate any help, thanks beforehand!

[@primaljeroen2405]

Idk if you still need it, but any group is closed under its operation.

[@primaljeroen2405]

@Hamza Zaidi I'm not sure if I know what you're asking me but I'll try to explain something. First try to understand that we're proving that the cosets act like a group. One property of groups is closure under its operation. To prove this, we must show that (xN)*(yN) is still a coset in which * is a random operation. I dont necessarily think multiplication is the only operation you can use here but I could be wrong about that. If you're confused about the notation (xN)(yN): this isn't necessarily multiplication, it's just an arbitrary operation noted this way because math is lazy. Hope this helped

[@primaljeroen2405]

@Hamza Zaidi yes

[@SpiritVector]

Simple groups are the primes of group theory.

[@pubudunuwan1751]

your all videos are very descriptive .it helps to solve many troubles .i wish to do more and more videos. thank you

[@ingun37]

What does “product of coset” means at 6:12 ? like cartesian product? but instead of pairs (x,y) you use x*y?

[@mike11022]

No, it's the group operation in the quotient group (the elements are the cosets, the identity is the normal subgroup). At 6:12, we show that this product of cosets are well defined because product (group operation of the original group) of any elements of two cosets belong to the same third coset. Then next it's trivial to show that the cosets under this operation form a group, i.e. the quotient group.

[@crigsbe]

Socratica teaches the mathematical language in a most interesting, enjoyable way. You only must apply mathematical formalism just as we learned finally at secondary school level, doing simple algebra. Trigonometry is just as simple when you discover that all is based on similarity. Just follow the rules ! Most surprisingly is to realise, that the Pythagorean theorem is staring at us. Just use the hight onto the hypothenuse and discover 3 similar triangles. The hypothenuse is represented as a sum. Now apply simple similarity proportion. Voilà !!! Groups were staring at us and we did not see ! Mathematicians are eye opener.

[@maulikjain3894]

one can listen this forever/..

[@njahnavi7943]

Thanks a ton !!! Explained with such clarity. It was to the point , excellent explanation.❤️

[@sudarshann7194]

Is it nityananda who's in your profile?? 🤔

[@Adam-r2n]

@@sudarshann7194 eh ktir excellent

[@KeystoneScience]

just in time for finals ;)

[@Socratica]

Hooray! That's what we were hoping. :D Good luck!!

[@CreolLanguag]

@@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?

[@haroonahmad1850]

@@CreolLanguag good question. What's the answer of this question? Did you get it?

[@jamaluddin9158]

@@CreolLanguag Yes that is correct!

[@rekarlopunzalan]

Watching this during finals

[@tsunghan_yu]

I'm lost from 6:05 :( and I also don't know how to do the exercise at the end of the video.

[@CrazyGamer-tw8fe]

i too didnt understand the condition

[@cameronspalding9792]

@10:45 if you find the factors associated with the composition series: is it possible to then reconstruct the group after factoring it

[@quahntasy]

This was explained very amazingly. Thanks for this :)

[@narendrakhadka9598]

You are my best teacher.

[@Socratica]

Sign up to our email list to be notified when we release more Abstract Algebra content: snu.socratica.com/abstract-algebra

[@murielfang755]

So so great. So well delivered.

[@souravsingh3025]

Great explaination love it , makes the topic fun 💝💝

[@undergraduateMath]

When I think of the 2-3hrs my lecturer spends in class, I just sit back and watch SOCRATICA... 3hours on here is......... You know what I mean.

[@moularaoul643]

Thank you so much!!!

[@theboombody]

This video was very handy for me.

[@LuizaDreamsDeutsch]

Thank you so much for the explanation!

[@sinisternightcore3489]

My solution for the exercise. Would like to confirm: I found 4 subgroups, 1 of order 3 and 3 of order 2. Only the one with 3 elements is a normal subgroup.

[@valeriobertoncello1809]

I think you're right, because noone of the order 2 subgroups can be treated as the neutral element of a quotient group.

[@karthik-jn3eb]

can someone please explain what happens at 6:10 and 6:11

[@j.sparks3211]

Groups must be closed under their associated binary operation; that is, for any elements a & b in G, a*b must also be in G (where * is the group G's binary operation). Think of the integers under addition (which is a group whose binary operation is addition). If you add any two arbitrary integers you get another integer. It's not until you introduce the operation of division (i.e. multiplication by an arbitrary integer's inverse) that you "step out of" the integers. Think about it, the inverse of 2 (under multiplication) is 1/2. 1/2 is not an integer, so by taking the multiplicative inverse of two, you get another set, namely, the Rational Numbers. This means that the Integers paired with the binary operation of multiplication is NOT a group, since for every integer there does not exist a multiplicative inverse in the integers. In other words, the Integers are NOT closed under division. So, back to the video. For the respective cosets of N, xN & yN, to behave like a group, they must be closed under the groups binary operation; i.e. any element in xN times any element in yN must be in (xN)(yN), i.e., (xN)(yN) must be closed under the groups binary operation. Did that help at all? If anyone is reading this spots a mistake, please do say. I'm currently learning this subject myself.

[@ueiwqoak]

@@j.sparks3211 I bet you're right but I'm still confused. When we're talking about closure for respective cosets of xN and yN, and xN, yN respectively being elements of the quotient group created by N, then doesn't that mean just that xN*yN is an element of the quotient group created by N? I don't get where we can say anything about x and y being elements of xN*yN. I would think we should be talking about xN*yN being an element of something itself. Using the definition of closure.

[@amberszulc197]

that was..... amazing. great job

[@user-dp9yn7zf4l]

6:22 i dont get y it wont equal to xyN y not xyN^2 can someone explain to me plz

[@MuffinsAPlenty]

The trick here is that N^2 = N. Why? Well, we have to understand what it means to multiply together _subsets_ of a group. If S and T are subsets of a group, we define the product ST to be the set {st | s is in S and t is in T}. In other words, we take all products of elements from S and T. Since N is a subgroup, N^2 = NN is actually equal to N itself! Why? First, let's check that N^2 is contained in N. Since N is a subgroup, N is closed under multiplication, so if s and t are both elements of N, then their product st is an element of N as well. So we get N^2 is contained in N. What about the reverse containment? Is N contained in N^2? Yes, and the reason why is that N has the identity element! Let e be the identity element. Then for any element s in N, s = se, which is an element of N^2.

[@zhiyizhu3040]

In 06:15, in the 4th line you wrote, for cosets to act like a group, x•y belongs to (xN)•(yN). Why is it true?

[@KGafterdark]

I got confused by this at first too, but it's fairly easy. You need to understand how a group multiplication over two SETS is defined. (xN)*(yN) means {xn*yn|x,y in G, n in N} i.e. the set of all possible pairwise combinations of (xn) multiplied by (yn). Since the identity element has to be in N, x has to be in xN and y has to be in yN because we multiply all elements of xN with all element of yN, we get that xy has to be in (xN)*(yN).

[@b43xoit]

@@KGafterdark Are you saying that for each pair in the Cartesian product of the two sets, you take the product of the elements of the pair under the group operation?

[@KGafterdark]

@@b43xoit Yes.

[@b43xoit]

@@KGafterdark Can you give an example of a group G having a subgroup H such that the left cosets generated by H in the context of G do not form a group?

The section at 6:05 expanded: (xN)(yN) = {x*n₁*y*n₂ | n₁,n₂∈N} where e∈N so (for n₁,n₂=e) we have (xN)(yN) ∋ (x*e)*(y*e) = x*y = x*y*e ∈ xyN. We have shown that x*y ∈ xyN ∩ (xN)(yN). Now, how to show that (xN)(yN) = xyN follows from that?

[@ctrlsys-exe]

Such a beautiful topic

[@physicslover9227]

Thanks a lot this videos series is very useful. It explains everything in a very simple way🙂🙏🏻🙏🏻.

[@malicksoumare370]

My favourite teacher

[@ajaykanwar7223]

*Love the way you teach*

[@ubersham]

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?).

[@homiramanuj]

9:52 So the Quotient Group and the factor group are same? Do they have the same meaning?

[@pedroamaral3935]

Thx for the video. The "important note" at 4:39. Everybody else understood that ?... How come "it's an entirely different group" than Z ?... What exactly is different ? A small constructive criticism: I do think it's easier to work with left coset of N == right coset of N, than the conjugation. At least to solve problems, like your homework assignment... And as someone else points out: y^-1.N.y == N, does not mean that the *same* n1 element is conjugated to n1. (but to some other n2 in N) But ok, maybe I was not paying enough attention. And yes... really interesting and really, really hot!

[@ClumpypooCP]

Its an entirely different group than Z because it is literally an entirely different group

[@seththepeacock]

At 6:05, why is that necessary for the cosets to behave like a group?

[@anirbanroy8488]

Normal subgroup of s3={(1,2,3),(e),(1,3,2)}or{(e),(1,2),(1,3)}

[@coreconceptclasses7494]

{(12),(13),(e)} that's not group because not closed under permutation composition so that can't be normal

[@shockline1]

Very helpful videos. I had to pause a lot to understand but worth it.