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.
@@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 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,
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.
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.
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 .
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...
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 ❤❤
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)
@@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?
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
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.
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
The curly braces are incorrectly positioned (time mark 3:23 out of 8:54).
4 роки тому
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
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.
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.
5:37 "If you take the direct product of infinitely many finite groups, you get an infinite group." Doesn't this also require that infinitely many of those groups are not the trivial group? Otherwise, you could just multiply the trivial group by itself and end up with a group that just has 1 element.
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
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?
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
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 :)
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!!!!!
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
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
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!
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.
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
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).
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.
Could you help clarify how to determine the number of elements of a certain order in direct product groups? Specifically, I am having trouble with the counting portion. Once I have all the possible order combinations for, say |a|, |b|, and |c| in the direct product of some ZnxZmxZp (direct product of three mods), where do I go from there? There seems to be a lot up on youtube near the question I am asking, but then no one explains how I count the number of elements, and the logic behind why I count it that way. I hope that makes sense.
I don't really get your question. Are you asking how do you know how many elements are in your group? Like the integers mod 5 under addition would be the set of 0,1,2,3,4 so the order of the group would be 5 And the integers mod 3 under addition would be the set 0,1,2 , with order 3 So |G| = 3x5 = 15 I really think I misunderstood your question, if you can clarify it a bit. Maybe I can attempt to answer it. Cause the order is just the number of elements in the group
(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
I can't wait for my baby niece to grow up and ace all her stem courses because Socratica exists
You're awesome
This video is awesome! Please do one on Sylow's theorems.
I binge watched 20 of her abstract algebra videos. It is just that good.
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 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.
You are an angel on Earth. You have no idea how many lives you're changing!
Amazing videos! I think I've already watched ~8 so far this morning, and I have no plans of stopping!
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.
this was an interesting and very well organized math lesson. thank you for the encouragement.
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 .
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...
Thank you for making this video brief . It helped me to learn the concept . ❤️. Take love from INDIA 🇮🇳
Getting back to brass tacks. Awesome.
Thanks!
Thank you so much for your kind support! We so love making these videos. 💜🦉
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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 should do a video on semi-direct products.
I feel like I was learning linguistic or grammar things! Interesting! Thank you!
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
At 4:55, ans is (2,2,4)
An infinite product of trivial groups is a trivial group. This was overlooked (time mark 5:37 out of 8:54).
well explained 👍.
can't wait for semi-direct product. so excited!
your method of teaching is amazing.. keep it up
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
The elements of the quotient group Z/3Z and of Z sub 3 are confused (time mark 4:38)
these are the best videos ever...thank you for that
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?
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.
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.
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.
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.
You are always special and make topics friendly!!!
"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.
this channel is my hero
Your explaination is sooo simple I like it and can understand it very much thank you Ma'am.
Can you help explain how you did the multiplication at time 6:11 (or point me to where it is explained)
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
It's an absolutely way to teach us and made us more effectively
@3:11, what's the answer?
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...
This is such a precise explanation. Thank you mam
8:03 simpler groups mean referring to cyclic groups not simple groups right!
Love from India you are great teacher ❤️
The curly braces are incorrectly positioned (time mark 3:23 out of 8:54).
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
It's very helpful, 💯
Thank you...
Additional !
I wish, I're also speak fluent English like you...
Wow, fantastic explanation
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.
THIS VIDEO IS AWESOME.THANK YOU!
Can you talk about semi-direct product?
plz
Make some vdos on external and internal direct product of group... ??
And what's the difference between them ??
Pls reply
Ingeneral these are same
Plz make more videos on EDP and complete modern algebra
well explained
8:41 So what is the inverse of a patron?
What even is the operation on the group?
Thank u so much can you explain the Fundamental
Theorem of Finite Abelian Groups
Exception to infinite number of finite groups: if all but a finite number of groups are the trivial group of the identity.
is all professor explain the way you do then only few would dislike math
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.
Helpful as always
Nice job keep it up
this is very helpful video.....thankyou so much
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
I love ur teaching
plz explain in detail... how a set transforms to a space & space to metric space with examples?
Which are the ways to piece together groups other than the direct product? i could not find those ways anywhere.
5:37 "If you take the direct product of infinitely many finite groups, you get an infinite group."
Doesn't this also require that infinitely many of those groups are not the trivial group? Otherwise, you could just multiply the trivial group by itself and end up with a group that just has 1 element.
Yes, you need infinitely many groups to be nontrivial.
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.
(12 , -I) inv (-12 , i)
Thank you so much.
Great video.
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?
Hey ur all videoes r vry useful to understand can u explain product of 2 subgroups i really need this plz
Great videos!
This is the best video
this video reminds me of PBS Infinite series 🤔
But it's more technical.
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
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)}.
Thanks
Thankyouuuu so much!!! 🙏🏻😍❤❤
Ecxellent lecture
you re wonderful. thank u thank u thank u..
I love the music ......
thank you
Sorry For my Question, But Y Would Somebody Thinks Of Making More Greater Or Complex Groups??!! N What Is The Aim Behind It..??
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!!!!!
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
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
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!
Thank you so much now i understand the concept.
Oh you are definitely wrong. We identify each direct factor independently with the canonical image of any 2 of them having a trivial intersection.
thank you madam.....
Done!
Amazing.
I watch you videos, they are great
but can create a video about Planar near-ring
thanks! really helpful:)
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
Subgroups of Z+Z(external direct product) ???
Mam will you find direct product product of A3 and S3 ? For me
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
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).
is this same as external direct product?
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.
Could you help clarify how to determine the number of elements of a certain order in direct product groups? Specifically, I am having trouble with the counting portion. Once I have all the possible order combinations for, say |a|, |b|, and |c| in the direct product of some ZnxZmxZp (direct product of three mods), where do I go from there? There seems to be a lot up on youtube near the question I am asking, but then no one explains how I count the number of elements, and the logic behind why I count it that way.
I hope that makes sense.
Also ... it would be really cool to hear how to find the number of both cyclic and non-cyclic subgroups of direct products. ^^;;
I don't really get your question. Are you asking how do you know how many elements are in your group?
Like the integers mod 5 under addition would be the set of 0,1,2,3,4 so the order of the group would be 5
And the integers mod 3 under addition would be the set 0,1,2 , with order 3
So |G| = 3x5 = 15
I really think I misunderstood your question, if you can clarify it a bit. Maybe I can attempt to answer it. Cause the order is just the number of elements in the group
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.