Direct Products of Groups (Abstract Algebra)
ฝัง
- เผยแพร่เมื่อ 13 ก.ค. 2024
- The direct product is a way to combine two groups into a new, larger group. Just as you can factor integers into prime numbers, you can break apart some groups into a direct product of simpler groups.
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Dummit & Foote, Abstract Algebra 3rd Edition
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Milne, Algebra Course Notes (available free online)
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(2,2,4) Btw. These are the best videos ever. As soon as I become a data scientist I won't forget who helped me!
How does group theory help with data science (just curious, cause I'm interested in both but didn't realize they relate)?
Karan Chopra Symmetries are a central concept in understanding the laws of nature, it is widely used in physics, mathematics, chemistry and machine learning.
an example of group symmetry being used in machine learning is Convolutional Neural Networks also known as space invariant artificial neural networks (SIANN). Mathematically, it is technically a sliding dot product or cross-correlation. This has significance for the indices in the matrix, in that it affects how weight is determined at a specific index point. Of course as soon as we are in the realm of matrices and linear algebra we are already being governed in many aspects by abstract algebraic structures and group theory.
same for me
@@LastvanLichtenGlorie translation invariance is cool but I feel like topological data analysis is much more of a direct application. At least from what I’ve seen
I don't think it's necessary to study this topic to become a data scientist
This video is awesome! Please do one on Sylow's theorems.
I can't wait for my baby niece to grow up and ace all her stem courses because Socratica exists
You're awesome
I hope this message finds you well. I wanted to take a moment to express my sincere gratitude for the exceptional lecture video you created. Your dedication to delivering valuable content and sharing your knowledge has not gone unnoticed, and I am truly thankful for the opportunity to learn from your video.
Your lecture was not only informative but also engaging and well-structured. The way you explained complex concepts with clarity and enthusiasm made the subject matter more accessible and enjoyable to grasp. Your expertise and passion for the topic shone through, making it a truly enlightening experience for me.
As a learner, I value educational resources that inspire and empower, and your video certainly did just that. Your commitment to fostering a deeper understanding of the subject is truly commendable.
Once again, thank you for your time and effort in creating such a valuable educational resource. Your dedication to sharing knowledge has undoubtedly made a positive impact on my learning journey. I look forward to exploring more of your content in the future.
With utmost appreciation,
What a cheering message to read. Thank you for your kind thoughts. We're so glad you've found us! 💜🦉
I binge watched 20 of her abstract algebra videos. It is just that good.
hey, I loved these series of videos on Group theory as we were generally told that its one of the most complex subjects but you have put it up really nicely.
Amazing videos! I think I've already watched ~8 so far this morning, and I have no plans of stopping!
this was an interesting and very well organized math lesson. thank you for the encouragement.
well explained 👍.
can't wait for semi-direct product. so excited!
I had no idea what you were talking about but here's a thumbs up anyway.
Due to a change of course I'm trying to study Abstract Algebra without having done the pre-module Geometry and Groups. These videos are a godsend.
Oh and due to my strange course through university, I've already got a 1st in Vector Spaces, a pre-requisite of which is Abstract Algebra.
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Pleaseeee more abstract algebra videos!! There are not enough videos and you make the best videos everrr!!!❤ you explain everything so perfect , i love u ❤❤
You are an angel on Earth. You have no idea how many lives you're changing!
Thank you for making this video brief . It helped me to learn the concept . ❤️. Take love from INDIA 🇮🇳
You are always special and make topics friendly!!!
I feel like I was learning linguistic or grammar things! Interesting! Thank you!
your method of teaching is amazing.. keep it up
Getting back to brass tacks. Awesome.
these are the best videos ever...thank you for that
Your explaination is sooo simple I like it and can understand it very much thank you Ma'am.
Awesome. Best way to define the product of two groups and their properties. Thank you.
"Direct" product. There are other kinds of products, like free products, and tensor products, semidirect product.
It would be lot useful if u even add the references at the end or beginning of the video ,content is elegantly explained in such short amount of time ,It would be lot useful if u can update videos on advance tops as well (ex:topology ,complex anylisis,real anylisis,number theory etc) .any way thanks alot it really helped me a lot to find the missing dots of my understanding in abstract algebra .
Thank u very much for ur contribution for us mam.u have just package of knowledge with good communication skill that directly touch my heart and it makes me productive.
Very good job. Thanks.
You should do a video on semi-direct products.
The rate at which videos are made here is direct product of subscribers, no. of viewers and Patreon supporters. Mathematical way!!!! Yeah!!! Loved the presentation...
This is such a precise explanation. Thank you mam
It's an absolutely way to teach us and made us more effectively
Wow, fantastic explanation
Im study masters degree as Electrical Egeneer and I have a course "deep learning and groups(groups, representation on groups, equivarints, symmetries) you just saved mee
Helpful as always
THIS VIDEO IS AWESOME.THANK YOU!
Great videos!
this channel is my hero
You are the best teacher of mathematical please make some videos for complex analysis
Nice job keep it up
It's very helpful, 💯
Thank you...
Additional !
I wish, I're also speak fluent English like you...
I wish I had a teacher like you back 25 years ago when I was in my pre-teens and in love with mathematics. it could've changed my life.
bad teachers year after year had me lose interest
let's not she's not alone in making this. Michael Harrison
and Kimberly Hatch Harrison also worked on this too.
Thank you so much.
Great video.
Thankyouuuu so much!!! 🙏🏻😍❤❤
thanks! really helpful:)
this is very helpful video.....thankyou so much
Another well explained video. Though I will say, a lot of this felt rather trivial, until the casual mention of that theorem at the end...
"At Socratica the rate at which we make videos is a *direct product* of views, subscribers and Patreon supporters"
She's lying. They intersect at the staff and equipment non-trivially so the product is internal and not direct.
I love ur teaching
well explained
Plz make more videos on EDP and complete modern algebra
Thank u so much can you explain the Fundamental
Theorem of Finite Abelian Groups
you re wonderful. thank u thank u thank u..
Hey ur all videoes r vry useful to understand can u explain product of 2 subgroups i really need this plz
Love from India you are great teacher ❤️
This is the best video
Done!
thank you
I think the correct law would be an infinite number of nontrivial finite groups (such that G is not equal to the set containing the identity element) have an infinite direct product
Yes this is by basic set theory.
There's also a direct sum that's restricted to abelian groups. They behave just like direct products except for in infinite number of summands, only finitely many components can be a non identity.
Amazing.
Can you talk about semi-direct product?
plz
Can I ask: if you break up a group into simple groups using the Jordan Holder theorem, how do you recombine them to get back to the original group
plz explain in detail... how a set transforms to a space & space to metric space with examples?
thank you madam.....
Ecxellent lecture
I watch you videos, they are great
but can create a video about Planar near-ring
Thanks
4:55 Is it (2,2,4)?
for a sec i thought it was (3,3,5) but we're using addition so e=0 and not 1 silly me!
True
I love the music ......
An infinite product of trivial groups is a trivial group. This was overlooked (time mark 5:37 out of 8:54).
i want ask about the weak direct product
I need to help thank you
Inspired by a comment about introducing lcm: I believe (n mod 3, n mod 5, n mod 6) for the same n would form a normal subgroup of 30 elements with cosets (n mod 3, n mod 5, (n+1) mod 6) and (n mod 3, n mod 5, (n+2) mod 6), and these three sets would form a quotient group, so that would say something new about the group Z/3Z x Z/5Z x Z/6Z, would that be right?
The elements of the quotient group Z/3Z and of Z sub 3 are confused (time mark 4:38)
You are great dear mathematician.. i wish to see some more topics in group theory like solvable group's , nilpotent group , normal series's and even module or galois theory too and we can make algebra playlist to more big
It's galois theory
@@Grassmpl yeah that what i mean. Thanks for correcting mine typo.
Can you help explain how you did the multiplication at time 6:11 (or point me to where it is explained)
Which are the ways to piece together groups other than the direct product? i could not find those ways anywhere.
please upload sylow theorem
4:56 : two inverses :
(-1, -3, -2) or (2, 2, 4)
In the group you're speaking of, (-1, -3, -2) = (2, 2, 4). :)
If you do it like that, you have infinitely many, but they are all identical.
I think you're forgetting (8, 7, 10) and (-7, -13, -14). She might as well give the short lecture on equivalence classes.
@@MuffinsAPlenty yes it was implicit in my mind, I should have written this ;)
@@jonmolina948 ((8,7,10) is the same as ( (2 = 8 / 3ℤ), (2 = 7 / 5ℤ), (4 = 10 / 6ℤ) ) so it seems that: ( (2 ≣ 2 + 3ℤ), (2 ≣ 2 + 5ℤ), (4 ≣ 4 + 6ℤ) ) then we can write ( ( (k ≣ k + 3ℤ), (l ≣ l + 5ℤ), (m ≣ m + 6ℤ) ) with k, l, m ∈ ℤ .
So all the inverses of (1, 3, 2) can be written as ( (2 + 3ℤ), ( 2 + 5ℤ), ( 4 + 6ℤ ) ) isn't it?
What are equivalence classes?
this video reminds me of PBS Infinite series 🤔
But it's more technical.
At 4:55, ans is (2,2,4)
So when we say you can factor groups like you can factor numbers into primes...
What we're really saying is that there exist an infinite number of unique groups, defined as the powers of some prime under multiplication. And that the direct product of this infinite number of groups is isomorphic to the rationals under multiplication.
Make some vdos on external and internal direct product of group... ??
And what's the difference between them ??
Pls reply
Ingeneral these are same
is all professor explain the way you do then only few would dislike math
Exception to infinite number of finite groups: if all but a finite number of groups are the trivial group of the identity.
You say there are no restrictions on the groups, but do they both have to be the same multiplication? For instance can G1 be mod20 and G2 = mod3?
Asked before I finished the video... So, to answer my own question... No, there can be different group operations when using the direct product :) Great videos, thank you!!!!!
The curly braces are incorrectly positioned (time mark 3:23 out of 8:54).
I'm confused. at 6:05(ish) you said a=(2,3) and b=(1,2)... don't all members of Sn have three parts? Where's the one in a and the three in b? And you multiplied two 2-tuples and got a 3-tuple, how's that work?
@Dana Hill
Yes, All members of Sn have 3 parts. The example uses something called "cycle notation." A couple of good youtubes on that - but in short the notation just follows where the number goes as you cycle through the elements in the parenthesis and then back to the first element. In other words [ a = (2 3) ] means a permutation where 2 goes to 3, 3 goes back to 2 with 1 staying fixed. b = (1 2) means 1 goes to 2 and 2 goes back to 1 while 3 stays fixed. So doing multiplication or rather the composition of a*b we just determine where a number ends up. So let's start with 1 in Sn =(1 ? ?) and then the second element (?) is where we end up. Do b = (1 2) first, so (1 goes to 2) and then to see where 2 goes we look at a = (2 3) and see (2 goes to 3) so a*b takes 1 to 3. In cycle notation we write 1 goes to 3 as (1 3 ?). Now let's see where 3 goes? b (3 stays fixed) and then a (3 goes to 2) so a*b takes 3 to 2 and we write that in the final as (1 3 2) -- (1 goes to 3 goes to 2)
Now for b*a (1 ? ? ): a leaves 1 fixed and b takes 1 to 2, so b*a takes 1 to 2 = (1 2 ?). then for the final element a takes 2 to 3 and b leaves 3 fixed so b*a takes 2 to 3, so we write (1 2 3) - (1 goes to 2 goes to 3)
ok
Thanks for that!
Maybe they should have a note about that, maybe a link to another video about it.
a keeps 1 fixed so it is omitted and b keeps 3 fixed so it is also omitted in o/w it would be a=(23)(1), b=(12)(3)
thank you for this thing
Thanks mam
5:30 Technically if you took a direct product of a finite number of finite groups and infinite number of groups of order 1 (with just a single element, namely identity element) you'd still get a finite group, right?
I know it's a corner case but still :D
Brilliant mam
extremely useful... tq so much... can u plz upload real analysis videos too? i wil b really useful plz plz plz
Know your limits buddy. Also practice your epsilon delta proofs.
At 5:37 the video says that if you take the direct product of an infinite number of finite groups, you get an infinite group. Surely this is only the case if an infinite number of those groups are non-trivial? Maybe that goes without saying, but I wanted to check. For example, if you took the direct product of the integers mod 3, with an infinite number of groups of order 1, you would get a group of order 3, right? Not an infinite group.
You are correct.
Thanks mam ,,,,
What do you mean by finitely generated?
It is really amazing video and I have learnt a lot. However, I have a question regarding the direct product of real numbers and symmetry group S3, why the a=(2 3) and b=(1 2) can operate under the multiplication of symmetry group S3 since it contain two elements inside a and b which is different from the previous symmetry group video that contain 3 elements? Thank you very much and I love Socratica video so much! Hope can come out more Abstract linear algebra video ! :)
@Chong (tried to explain one way to Dana Hill above will try another way with you - hopefully you both see and at least one makes sense to either one - maybe both :))
a = (2 3) is cycle notation and just means element 2 goes to element 3, element 3 loops back to element 2, and element 1 stays fixed. So (x y z) would go to (x z y) or using numbers (1 2 3) goes to (1 3 2)
b = (1 2) means 1 goes to 2, 2 goes back to 1 and 3 stays fixed. (x y z) goes to (y x z) or (1 2 3) goes to (2 1 3)
a*b means b first (x y z) goes to (y x z) then a takes (y x z) to (y z x) or (xyz) goes to (yzx) which if we write that in cycle notation (1 3 2) = element 1 goes to element 3, element 3 goes to element 2, element 2 goes to element 1
b*a means a first (x y z) goes to (x z y), then b takes (x z y) to (z x y) or (xyz) goes to (zxy). In cycle notation (1 2 3)
Again the (2 3) and (1 2) are not elements of the Sn group, just a way to write how the permutation happens - what it does. Just like the products (1 2 3) and (1 3 2) are not elements of Sn but are just how the permutation happens (1 2 3) = element 1 moves to the element 2 spot, element 2 moves to the element 3 spot and the element 3 spot loops back to the element 1 spot.
The product of (2 3)and (1 2) is nothing but the product of two cycle
5:30 if we take a product of infinite trivial groups we get a trivial group, which is finite. So this statement is not always true.
Sorry For my Question, But Y Would Somebody Thinks Of Making More Greater Or Complex Groups??!! N What Is The Aim Behind It..??
Mam will you find direct product product of A3 and S3 ? For me
Okay mam can you tell us Cn×Cm is isomorphic to Cnm
Can I use induction to show that dp will be non abelian, if any of the factor is non abelians
That won't work for the infinite case. Even restricting to a finite case wouldn't that be an overkill?
I don't get it. Lets say we have two groups A = { Z/2 (integers mod 2) } and B = { Z/4 (integers mod 4) }. The groups have different order |A| = 2 and |B| = 4 so if we calculate C = A x B, C would have order 8, but what would the elements of C be? Step by step answer would be appreciated :)
C = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}.
(12 , -I) inv (-12 , i)