Direct Products of Groups (Abstract Algebra)

[@LastvanLichtenGlorie]

(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!

[@KCIsMe]

How does group theory help with data science (just curious, cause I'm interested in both but didn't realize they relate)?

[@LastvanLichtenGlorie]

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.

[@geogeo14000]

same for me

[@chymoney1]

@@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

[@azhari7968]

I don't think it's necessary to study this topic to become a data scientist

[@ninosawbrzostowiecki1892]

This video is awesome! Please do one on Sylow's theorems.

[@almustaphaumar2904]

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,

[@Socratica]

What a cheering message to read. Thank you for your kind thoughts. We're so glad you've found us! 💜🦉

[@m06prc]

I binge watched 20 of her abstract algebra videos. It is just that good.

[@coldblaze100]

I can't wait for my baby niece to grow up and ace all her stem courses because Socratica exists

[@ecowithsudheer5547]

You're awesome

[@bencrossley647]

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.

[@thematt5325]

I had no idea what you were talking about but here's a thumbs up anyway.

[@PunmasterSTP]

Amazing videos! I think I've already watched ~8 so far this morning, and I have no plans of stopping!

[@Fatim15leo]

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.

[@saiganeshreddy7235]

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 .

[@rayrocher6887]

this was an interesting and very well organized math lesson. thank you for the encouragement.

[@parimuffins]

You are an angel on Earth. You have no idea how many lives you're changing!

[@categorygrp]

You should do a video on semi-direct products.

[@francceD]

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 ❤❤

[@amitakar9620]

Thank you for making this video brief . It helped me to learn the concept . ❤️. Take love from INDIA 🇮🇳

[@sanjityadav3772]

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...

[@jpphoton]

Getting back to brass tacks. Awesome.

[@utomato1147]

I feel like I was learning linguistic or grammar things! Interesting! Thank you!

[@hasanaljamea2569]

well explained 👍. can't wait for semi-direct product. so excited!

[@sanan22]

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

[@jonm8218]

let's not she's not alone in making this. ​Michael​ ​Harrison and Kimberly​ ​Hatch​ ​Harrison also worked on this too.

[@fahrenheit2101]

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...

[@labmusqooraf5406]

your method of teaching is amazing.. keep it up

[@shravandanga1403]

Awesome. Best way to define the product of two groups and their properties. Thank you.

[@Grassmpl]

"Direct" product. There are other kinds of products, like free products, and tensor products, semidirect product.

[@יעקבמישייב]

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

[@markmathman]

An infinite product of trivial groups is a trivial group. This was overlooked (time mark 5:37 out of 8:54).

[@amritathakur2036]

Plz make more videos on EDP and complete modern algebra

[@RAJSINGH-of9iy]

At 4:55, ans is (2,2,4)

[@AdityaKumar-bv3hq]

Wow, fantastic explanation

[@guilhemescudero9114]

4:56 : two inverses : (-1, -3, -2) or (2, 2, 4)

[@MuffinsAPlenty]

In the group you're speaking of, (-1, -3, -2) = (2, 2, 4). :)

[@badhbhchadh]

If you do it like that, you have infinitely many, but they are all identical.

[@jonmolina948]

I think you're forgetting (8, 7, 10) and (-7, -13, -14). She might as well give the short lecture on equivalence classes.

[@guilhemescudero9114]

​@@MuffinsAPlenty yes it was implicit in my mind, I should have written this ;)

[@guilhemescudero9114]

​@@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?

[@paominlienguite2397]

Your explaination is sooo simple I like it and can understand it very much thank you Ma'am.

[@1729STUDYPOINT]

It's very helpful, 💯 Thank you... Additional ! I wish, I're also speak fluent English like you...

[@ScilexGuitar]

4:55 Is it (2,2,4)?

[@alberttresvilanova6953]

for a sec i thought it was (3,3,5) but we're using addition so e=0 and not 1 silly me!

[@durastory1928]

True

[@DanaLea57]

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?

[@kellyepperson6168]

@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)

[@DanaLea57]

ok

[@missingno9]

Thanks for that! Maybe they should have a note about that, maybe a link to another video about it.

[@RoyalH3iR]

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)

[@avishkadhananjaya6964]

thank you for this thing

[@rollbacked]

this channel is my hero

[@Kisnaram_r_j]

It's an absolutely way to teach us and made us more effectively

[@arbsieyasin4127]

You are always special and make topics friendly!!!

[@sarojpadhy5535]

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.

[@nareshbarik5384]

thank you

[@avishkadhananjaya6964]

these are the best videos ever...thank you for that

[@mathsacademy4174]

well explained

[@huttarl]

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.

[@MuffinsAPlenty]

You are correct.

[@poulamikarmakar3759]

Make some vdos on external and internal direct product of group... ?? And what's the difference between them ?? Pls reply

[@alishasahraribaarisha4740]

Ingeneral these are same

[@markmathman]

The elements of the quotient group Z/3Z and of Z sub 3 are confused (time mark 4:38)

[@maryamsaeed3633]

This is such a precise explanation. Thank you mam

[@vaskoalph1648]

Love from India you are great teacher ❤️

[@munirahabdullah6542]

Thank u so much can you explain the Fundamental Theorem of Finite Abelian Groups

[@RurczakKurczak]

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.

[@kz.0681]

THIS VIDEO IS AWESOME.THANK YOU!

[@elielx2]

"At Socratica the rate at which we make videos is a *direct product* of views, subscribers and Patreon supporters"

[@Grassmpl]

She's lying. They intersect at the staff and equipment non-trivially so the product is internal and not direct.

[@bobbiemarkwick4417]

Exception to infinite number of finite groups: if all but a finite number of groups are the trivial group of the identity.

[@簡昱睿-n2x]

Can you talk about semi-direct product? plz

[@eddy1537]

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 :)

[@badhbhchadh]

C = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}.

[@GelidGanef]

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.

[@AbhishekSingh-mh4vk]

this is very helpful video.....thankyou so much

[@jayashree.j1777]

plz explain in detail... how a set transforms to a space & space to metric space with examples?

[@dipsacedemy]

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

[@Grassmpl]

It's galois theory

[@dipsacedemy]

@@Grassmpl yeah that what i mean. Thanks for correcting mine typo.

[@bigstroker1300]

Which are the ways to piece together groups other than the direct product? i could not find those ways anywhere.

[@Matchless_gift]

Helpful as always

[@cameronspalding9792]

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

[@MarkLares]

Can you help explain how you did the multiplication at time 6:11 (or point me to where it is explained)

[@mehulkumar3469]

When you take the direct product of integer mod (3×5×6) the order of the direct product should be lcm of all groups combined since all groups are cyclic hence i think the order of G must be 30 instead of 90 correct me if I am wrong

[@descuddlebat]

This would be true if it were (n mod 3, n mod 5, n mod 6) for the same n, but that is not a requirement. As it is, your choices for each element are independent - you have 3 choices for first element to be an integer mod 3, and similarly 5 choices for second element and 6 for third, leaving you with 3x5x6 = 90 total choices

[@descuddlebat]

That said, I believe (n mod 3, n mod 5, n mod 6) 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 your idea with employing lcm actually says something new about the group!

[@mehulkumar3469]

Thank you so much now i understand the concept.

[@Grassmpl]

Oh you are definitely wrong. We identify each direct factor independently with the canonical image of any 2 of them having a trivial intersection.

[@descuddlebat]

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?

[@ramonalejandrobarajaslopez2482]

Hi. I am an English student and unfortunately I have a hearing impairment and I need to hear more the language to familiarize myself with, so taking into account that your videos are short, I would like to ask you to please put subtitles to your videos or I do not know if it is possible but Allow someone to put them on. I would greatly appreciate it. Greetings to all of you.

[@ramonalejandrobarajaslopez2482]

I'm a new sub.

[@matteolombardo1]

They are already subtitled...

[@ramonalejandrobarajaslopez2482]

They're generated automatically, they're not correct at all. Only few videos are captioned manually.

[@patrickrossi75]

Because she speaks so clearly the automatic subtitles are pretty accurate though in this case

[@sureshmathemetics6696]

I love ur teaching

[@Bharat_Rohan]

8:03 simpler groups mean referring to cyclic groups not simple groups right!

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

[@phooiseanchong1473]

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 ! :)

[@kellyepperson6168]

@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.

[@bhagabansharma5433]

The product of (2 3)and (1 2) is nothing but the product of two cycle

[@kunslipper]

Thank you so much.

[@HDQuote]

this video reminds me of PBS Infinite series 🤔

[@hafizajiaziz8773]

But it's more technical.

[@lipshabhoi6300]

Hey ur all videoes r vry useful to understand can u explain product of 2 subgroups i really need this plz

[@khushisang5744]

Answers of the questions are =>Inverse of (12,i) is (1,-i) =>No, every element doesn't have an inverse =>Associative property hold for this example M i right.. If no then tell me the true one

[@sandeepkumarvermaau3741]

Nice job keep it up

[@markmathman]

The curly braces are incorrectly positioned (time mark 3:23 out of 8:54).

[@macmos1]

Great video.

[@markmathman]

Great videos!

[@michaellangan5991]

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?

[@michaellangan5991]

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!!!!!

[@nowornever5598]

Thanks

[@saurabhsingh-ow7ue]

thank you madam.....

[@mahanthivenkataramana157]

Ecxellent lecture

[@voiceofnation7493]

Sorry For my Question, But Y Would Somebody Thinks Of Making More Greater Or Complex Groups??!! N What Is The Aim Behind It..??

[@Socratica]

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

[@sirluoyi2853]

Done!

[@tarigmergani3903]

is all professor explain the way you do then only few would dislike math

[@dhruvprajapati9036]

This is the best video

[@maikellaishram7484]

Subgroups of Z+Z(external direct product) ???

[@khushisang5744]

Answers of the questions are =>Inverse of (12,i) is (1,-i) =>Yes every element have an inverse =>Associative property hold for this example M i right.. If no then tell me the true one

[@cameronmyron5776]

Yes to 2 and 3, no to the first. The inverse of the element (12,i) in the group of the integers under addition X {1,-1,i,-i} under multiplication is (-12,-i) because 12+(-12)=0 and i*(-i)=1 (and the identity element is (0,1)). (12,i)*(1,-i)=(13,1) not (0,1).

[@mathonline6013]

Mam will you find direct product product of A3 and S3 ? For me

[@phyziks878]

(12 , -I) inv (-12 , i)

[@shelymutiaramaghfira]

Thankyouuuu so much!!! 🙏🏻😍❤❤

[@meenakshisingh4195]

please upload sylow theorem

[@mrgd7813]

you re wonderful. thank u thank u thank u..

[@zakirullahbzt]

Okay mam can you tell us Cn×Cm is isomorphic to Cnm

[@heathernapthine8775]

Why when we take the product G=R x S_3 do we add our elements in R but multiply those in S_3? Thanks

[@MuffinsAPlenty]

The group G = R x S_3 is built up from two smaller groups, R and S_3. The operations are done component-wise using the two individual groups. The first component uses the group R. What is the group operation on R? The second component uses the group S_3. What is the group operation on S_3?

[@fahrenheit2101]

​@@MuffinsAPlentywhat is the group operation on R? Depends... There are additive reals, but also multiplicative reals. Ideally one should specify. Though I suppose that since 0 wasnt excluded you can assume its additive.

[@MuffinsAPlenty]

@@fahrenheit2101 Yes! One of the things that tripped me up for a while when I first learned abstract algebra and topology is that, despite the fact that any set could have multiple structures placed on it, many sets we commonly use only have a unique _standard_ operation which we default to, unless otherwise specified. For the set of real numbers, you can come up with infinitely many operations which form a group on R. However, there are only two _standard_ binary operations on R: addition and multiplication. And you're exactly right that multiplication is _not_ a group operation on R because 0 is an element of R. So we only have one standard operation left: addition. So if someone mentions the set of real numbers as a group, and they don't specify an operation, everyone should assume the operation is addition. Similarly, if someone were to mention R\{0} as a group, one sees that addition is no longer a group operation, but multiplication is a group operation. So if someone mentions R\{0} as a group without specifying an operation, everyone should assume it the operation is multiplication. If someone wants to use a nonstandard group operation on the set of real numbers or on the set of nonzero real numbers, the onus is on them to specify the operation.

[@fahrenheit2101]

@@MuffinsAPlenty ah, thanks for the info! I didnt realise how commonplace these conventions were... (I'm new to Group Theory, currently studying my first course on it this term and it's fascinating, but also really weird Speaking of, I should catch up on the Groups lecture I skived today...)

[@randomdude9135]

What do you mean by finitely generated?

[@jayashree.j1777]

extremely useful... tq so much... can u plz upload real analysis videos too? i wil b really useful plz plz plz

[@Grassmpl]

Know your limits buddy. Also practice your epsilon delta proofs.

[@SwrwngThai]

I love the music ......

[@hayderkanaan3012]

I watch you videos, they are great but can create a video about Planar near-ring

[@vanguard7674]

Amazing.

[@ammar84ful]

is this same as external direct product?

[@Grassmpl]

Yes. There is also internal direct product: we need each factor to be a subgroup of a common larger group G. The elements here are element of G obtained from multiplying out the coordinates rather than keeping tuples. Furthermore, we must obtain a subgroup of G isomorphic to their external direct product.

[@zxdzxdzxd]

Can anyone tell me, why a=(2 3) b=(1 2) a*b=(1 3 2)? or in which video can i find this explaination.

[@Grassmpl]

It's a composition of permutations in cycle notation. 1-->2-->3 3 to 3 to 2 2 to 1 to 1 We get (1 3 2) erasing the middle numbers of each row.