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.
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.
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!
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.
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.
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 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.
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!
@@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
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)
@@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 :)
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!
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).
@@_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.
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 :)
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.
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 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)
@@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?
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.
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)
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.
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!
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
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.
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)
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.
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
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.
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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.
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.
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...
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 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?
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 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 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.
Great lecture! Thank you so much for presenting Abstract Algebra so eloquently. But I do not seem to understand one tiny bit. Why are Quotient Groups not a Subgroup? They seem to contain all the elements of G, in this case, all the integers Z. Which elements are missing in G/N?
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
Great lecture by the way! But I didn't understand why proving (xN)(yN)=xyN is sufficient to show the cosets form a Group? Edit: After couple of days of thinking and researching. I found that we make an assumption that (xN)(yN)=(xy)N. We can show that if (xN)(yN)=(zN) where z=xy, then the cosets form a group, it is clear the definition you multiply one coset by another coset and you get some another coset. The complete proof of proving it is a group is well described in the video. If (xN)(yN)=(xy)N is true for non-abelian group, then while proving LHS=RHS we have to make an assumption yN=Ny. If yN=Ny is true for all y, we say that the subgroup N is normal. Or we can rephrase it by multiplying by right inverse of y to both sides. yNy^-1=N. Therefore, Normal subgroup implies cosets form a group and vice versa.
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
@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
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?).
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
Thanks a lot this videos series is very useful. It explains everything in a very simple way🙂🙏🏻🙏🏻.
5 лет назад+1
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?
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.
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.
5:32 the music starting here is so beautiful, i cannot focus on the video :)) , i just close my eyes and listen the music, it is so relaxing, anyone knows what that song is?
If you'd like to learn more, we have a free course on Group Theory! www.socratica.com/courses/group-theory
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?
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!
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.
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.
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.
The way you simplify a complex concept is great!
The way she teaches and explains , totally incredible...!
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!
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....
@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
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
@@homiramanuj You always go for a number
The most understandable videos of abstract algebra on RUclips.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
@@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 :)
This playlist is awesome!
TOPOLOGY WHEN?!?!? 😁👍🏻
S... waiting
Have they uploaded it now?
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
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 most approachable abstract algebra course online. thank you so much for your hard work!
What an amazing series, this is a goldmine! Perfect depth and speed.
I use this series to accompany my lectures on abstract algebra, it helps me so much to understand what is going on. Thankyou!
Trying to find a word that describes my gratefulness for such incredible explanative videos.
she saved my whole fxxking life during the senior this fall
Excellent presentation: clear, to-the-point, fluid.
This really helped me understanding this topic. I was really confused and now all the confusion is gone.
Thanks a lot .
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).
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
The best video i ever viewed on youtube about group theory. Thanks alot
I am studying for my Algebra exam and these videos give an amazing extra insight and perspective on the matter. Thank you!
We're so glad you're finding our videos helpful! Good luck with your exam!! 💜🦉
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
Mam your way of delivering lecturer is amazing,outstanding.. God bless you
Was self studying Galois Theory and this helped to recap a lot of forgotten theorems, thanks a lot!
@7:25 replace y with y^(-1)
These videos are so helpful it's unreal
In really like the presentation style. Everything is very clear and all the explanations are easy to follow. Thank you so much
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 !
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.
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.
Too good explanation which covers most important part of normal subgroup. U are truly a good teacher.
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)
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!
Simple groups are the primes of group theory.
Thanks! Had to watch in 0.5x the speed to hang on, but very helpful! :)
Excellent work. Students are recommended to watch this video. It will help to motivate you properly.
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
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.
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)
studying for my math classes is enjoyable with Socratica
took me watching it twice to understand perfectly(have to oil my brain)....awesome to the point explanation.
This is our favourite thing about RUclips compared to classes - you can just rewatch! Thanks for sticking it out with us! 💜🦉
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.
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!
So well explained!!!! Thank you! I have an exam tomorrow. I have now more confidence than apprehension XD
You helped me so much! Thank you for these videos because my professor is not good.
I really like you because explain the subject in an easy and understandable way.
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
I wish I could upvote this video 100 times.
wow you took such a complicated subject and make it so simple.
Thanks!
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
You are so kind to send support for our videos, thank you so much! 💜🦉
Outstanding video lecture. This video lecture is very helpful for self-study.
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.
@11:03 the set {I, (123), (132)} is a subgroup of S3
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)
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.
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
Happiness is nothing but understanding stg properly. These vids series are fuckin' awesome.
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.
I think you're right, because noone of the order 2 subgroups can be treated as the neutral element of a quotient group.
WOW, so nice and easier to understand. Beats the textbook 100%.
Indian here really enjoying this series happy republic day guys
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
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.
your all videos are very descriptive .it helps to solve many troubles .i wish to do more and more videos. thank you
This was explained very amazingly. Thanks for this :)
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
one can listen this forever/..
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Abstract Algebra is so difficult :\ I hope this math minor is worth the work lol
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.
Thank you. Excellent presentation.
I am eagerly waiting for Real numbers, Complex Numbers, Probability , Number Theory etc
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.
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...
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 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?
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.
You have made things simple!....Thankyou
Normal subgroup of s3={(1,2,3),(e),(1,3,2)}or{(e),(1,2),(1,3)}
{(12),(13),(e)} that's not group because not closed under permutation composition so that can't be normal
I'm lost from 6:05 :( and I also don't know how to do the exercise at the end of the video.
i too didnt understand the condition
Great lecture! Thank you so much for presenting Abstract Algebra so eloquently. But I do not seem to understand one tiny bit. Why are Quotient Groups not a Subgroup? They seem to contain all the elements of G, in this case, all the integers Z. Which elements are missing in G/N?
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
Can u explain coset multiplication briefly
Great lecture by the way! But I didn't understand why proving (xN)(yN)=xyN is sufficient to show the cosets form a Group?
Edit: After couple of days of thinking and researching. I found that we make an assumption that (xN)(yN)=(xy)N. We can show that if (xN)(yN)=(zN) where z=xy, then the cosets form a group, it is clear the definition you multiply one coset by another coset and you get some another coset. The complete proof of proving it is a group is well described in the video.
If (xN)(yN)=(xy)N is true for non-abelian group, then while proving LHS=RHS we have to make an assumption yN=Ny. If yN=Ny is true for all y, we say that the subgroup N is normal. Or we can rephrase it by multiplying by right inverse of y to both sides. yNy^-1=N.
Therefore, Normal subgroup implies cosets form a group and vice versa.
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
Why do x^{-1} and y^{-1} exist at 6:55? Cosets do not have inverse elements in general, do they?
Hmm... Both x and y are contained in G, so they must have inverses.
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!
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
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?).
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
You are my best teacher.
Where did the series introduce symmetric groups, and particularly S sub 3?
This one? ruclips.net/video/3aNeCWRjh8I/видео.html
My favourite teacher
Such a beautiful topic
@10:45 if you find the factors associated with the composition series: is it possible to then reconstruct the group after factoring it
Great explaination love it , makes the topic fun 💝💝
So so great. So well delivered.
Thanks a lot this videos series is very useful. It explains everything in a very simple way🙂🙏🏻🙏🏻.
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?
What does “product of coset” means at 6:12 ? like cartesian product? but instead of pairs (x,y) you use x*y?
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.
I understood this so much better than Fraleigh's book. Thank you :)
6:22
i dont get y it wont equal to xyN
y not xyN^2
can someone explain to me plz
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.
At 6:05, why is that necessary for the cosets to behave like a group?
5:32 the music starting here is so beautiful, i cannot focus on the video :)) , i just close my eyes and listen the music, it is so relaxing, anyone knows what that song is?