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

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!

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.

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.

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....

“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?

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

S... waiting

Have they uploaded it now?

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

@@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 :)

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

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

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

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

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

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

@@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)

@@ThePharphis Thanks for taking the time to respond!

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

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.

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

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

@@sudarshann7194 eh ktir excellent

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

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

Why care that cosets form a group?

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!

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

@@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.

@@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)

but i have a more fundamental problem, I don;t understand what does the statement _ xN combined with yN = (xy)N even mean where the hell did the set xyN came from

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

Coursera has a good courses on Complex analysis, and some probability courses too, oh yeah and one on Galois Theory but that one will be very very hard because its taught by a Russian lady. For other courses you can look on MIT open courses, maybe you will find something.

It's because the addition/multiplication operation is now different. And the elements of the quotient group are the cosets of the original group, not single elements. For example, Z/5Z is not a subgroup of Z in the traditional sense. In the original group, 3+3=6 while in the quotient group, 3'+3'=1' (note that these are congruence classes, not numbers, so while 6 is an element in 1', it is not equal to it). On the other hand, 5Z is clearly a subgroup of Z. It's just a matter of strictly following the definition of a subgroup.

Just a recap to what @Aalap Shah has said, Quotient group is a set of subsets of the group. To be a subset of the group, a set should contains some of its elements or being empty. Therefore, quotient group is not a subset and by extension, not a subgroup

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.

I know its difficult but its motivating to see it in action. See on google how groups are used in cryptography and quantum physics. I find applications to make it more interesting and easy to remember concepts.

This is actually the easy part of Abstract Algebra .... ha-ha, just kidding

What's great is that your math minor will make the rest of life seem rather easy.

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

i too didnt understand the condition

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.

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

@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

@Hamza Zaidi yes

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!

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

@@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.

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

@@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.