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@sadiqurrahman2
5 жыл бұрын
You explained a confusing topic in the most easiest manner. Thanks a lot.
@zy9662
3 жыл бұрын
I'm still confused as to why she says that every element has an inverse. Is this a consequence of the suppositions or an axiom?
@shreyrao8119
3 жыл бұрын
@@zy9662 Hi, Every element has its own inverse as this is one of the conditions which needs to be met for a set to be classified as a group
@zy9662
3 жыл бұрын
@@shreyrao8119 OK so it's an axiom. Was confusing because the next property she showed (that each element appears exactly once in each column or row) was a consequence and not an axiom
@brianbutler2481
3 жыл бұрын
@@zy9662 In the definition of a group, every element has an inverse under the given operation. That fact is not a consequence of anything, just a property of groups.
@zy9662
3 жыл бұрын
@@brianbutler2481 i think your choosing of words is a bit sloppy, a property can be just a consequence of something, in particular the axioms. For example, the not finiteness of the primes, that's a property, and also a consequence of the definition of a prime number. So properties can be either consequences of axioms or axioms themselves.
@mehulkumar3469
4 жыл бұрын
The time when you say Cayley table somewhat like to solve a sudoku you win my heart. By the way, you are a good teacher.
@MoayyadYaghi
3 жыл бұрын
I literally went from Struggling in my abstract algebra course to actually loving it !! All love and support from Jordan.
@Socratica
3 жыл бұрын
This is so wonderful to hear - thank you for writing and letting us know! It really inspires us to keep going!! 💜🦉
@tristanreid
4 жыл бұрын
If anyone else is attempting to find the cayley tables, as assigned at the end: If you take a spreadsheet it makes it really easy. :) Also: she says that 3 of them are really the same. This part is pretty abstract, but what I think this means is that all the symbols are arbitrary, so you can switch 'a' and 'b' and it's really the same table. The only one that's really different (SPOILER ALERT!) is the one where you get the identity element by multiplying an element by itself (a^2 = E, b^2 = E, c^=E).
@dunisanisambo9946
3 жыл бұрын
She says that there are 2 distinct groups because 1 is abelian and the rest of them are normal groups.
@rajeevgodse2896
3 жыл бұрын
@@dunisanisambo9946 Actually, all of the groups are abelian! The smallest non-abelian group is the dihedral group of order 6.
@jonpritzker3314
2 жыл бұрын
Your comment helped me without spoiling the fun :)
@fahrenheit2101
Жыл бұрын
@@rajeevgodse2896 Really, I thought I found one of order 5... All elements self inverse, the rest fills itself in. table (only the interior): e a b c d a e c d b b d e a c c b d e a d c a b e What have I missed?
@fahrenheit2101
Жыл бұрын
@@rajeevgodse2896 Nevermind, turns out I needed to check associativity - I'm surprised that isn't a given.
@waynelast1685
4 жыл бұрын
at 4:10 when she says "e times a" she means "e operating on a" so it could be addition or multiplication ( or even some other operation not discussed so far in this series)
@jeovanny1976andres
3 жыл бұрын
She says actually a times e, but here order it's important. And yes you are allright.
@kirstens1389
8 жыл бұрын
These videos are really extremely helpful - too good to be true - for learning overall concepts.
@kingston9582
5 жыл бұрын
This lesson saved my life omg. Thank you so much for being thorough with this stuff, my professor was so vague!
@TheFhdude
5 жыл бұрын
Honestly, I watched many videos and read books to really grasp Groups but this presentation is the best hands down. It demystifies Groups and helps to understand it way better. Many thanks!
@randomdude9135
5 жыл бұрын
But how do you know that the associative law holds?
@jonatangarcia8564
5 жыл бұрын
@@randomdude9135 That's the definition of a group, that associative law holds. Now, if you take a concrete set, you have to prove that is a group (Proving that associative law holds).
@randomdude9135
5 жыл бұрын
@@jonatangarcia8564 Yeah how do you prove that the cayley table made by following the rules said by her always follows the associative law?
@jonatangarcia8564
5 жыл бұрын
@@randomdude9135 Cayley Tables are defined using a group, then, associative laws hold, because, since you use a group, and you use the elements of the group and use the same operation of the group, it holds. It's by definition of a Group
@SaebaRyo21
7 жыл бұрын
This really helped me because application of caley's table is useful in spectroscopy in chemistry. Symmetric Elements are arranged exactly like this and then we have to find the multiplication. Thanks Socratica for helping once again ^^
@youtwothirtyfive
2 жыл бұрын
These abstract algebra videos are extremely approachable and a lot of fun to watch. I'm really enjoying this series, especially this video! I worked through the exercise at the end and felt great when I got all four tables. Thank you!
@sandeepk4339
5 жыл бұрын
I'm from India, your explanation was outstanding.
@efeuzel1399
4 жыл бұрын
I am watching and liking this in 2020!
@markpetersenycong8723
4 жыл бұрын
Guess we are here because of online class due to the Covid-19 😂
@halilibrahimcetin9448
4 жыл бұрын
Been to math village in Turkey?
@sukhavaho
3 жыл бұрын
@@halilibrahimcetin9448 wow - that is cool! will they make you find the prime factors of some random large number before they let you in? (İyi tatiller, BTW!)
@into__the__wild5696
Жыл бұрын
i am in2023
@АялаБақытбек
Жыл бұрын
2023...
@fg_arnold
5 жыл бұрын
love the Gilliam / Python allusions at the end. good work Harrisons, as usual.
@ozzyfromspace
4 жыл бұрын
I kid you not, I used to generate these exact puzzles for myself (well, mine were slightly more broad because I never forced associativity) so it's so good to finally put a name to it: *Group Multiplication Tables.* I used to post questions about this on StackExchange under the name McMath and remember writing algorithms to solve these puzzles in college (before I dropped out lol). I wish I knew abstract algebra existed back then. Liliana de Castro and Team, at Socratica, you're phenomenal!
@JJ_TheGreat
5 жыл бұрын
This reminds me of Sudoku! :-)
@mheermance
5 жыл бұрын
I was just thinking "hey we're playing Sudoku!" when Liliana mentioned it at 6:30. As for the challenge. The integers under addition are the obvious first candidate, but the second unique table eluded me. I tried Grey code, but no luck, then I tried the integers with XOR and that seemed to work and produce a unique table.
@MrCEO-jw1vm
8 күн бұрын
couldn't hold my excitemnet and just kept saying "wow, wow"! I have found a new love subject in math. I'll take this class this fall!!! Thanks so much for this content. It has blessed my life!
@JozuaSijsling
4 жыл бұрын
Awesome video, well done as always. One thing that confused me was that group "multiplication" tables actually don't necessarily represent multiplication. Such as when |G|=3 the Cayley table actually represents an addition table rather than a multiplication table. I tend to get confused when terms overlap, luckily that doesn't happen too often.
@tomasito_2021
3 жыл бұрын
I have loved abstract algebra from the first time I read of it. Google describes it as a difficult topic in math but thanks to Socratica, I'm looking at Abstract algebra from a different view. Thanks Socratica
@hansteam
7 жыл бұрын
Thank you for these videos. I just started exploring abstract algebra and I'm glad I found this series. You make the subject much more approachable than I expected. The groups of order 4 was a fun exercise. Thanks for the tip on the duplicates :) Subscribed and supported. Thank you!
@yvanbrunel9734
4 жыл бұрын
the weird thing is I have to convince myself that "+" doesn't mean "plus" anymore 😩
@Abhishek._bombay
5 ай бұрын
Addition modulo 🙌😂
@arrpit5774
Жыл бұрын
Just loved your content , getting easier with each passing minute
@fahrenheit2101
Жыл бұрын
I've got the 2 groups - spoilers below: Alright, so they're both abelian, and you can quickly work them out by considering inverses. There are 3 non identity elements - call them *a*, *b* and *c*. Note that these names are just for clarity, and interchanging letters still keeps groups the same, so what matters isn't the specific letters, but how they relate. One option is to have all 3 elements be their own inverse i.e. *a^2 = b^2 = c^2 = e* Alternatively, you could have some element *a* be the inverse of *b*, and vice versa, such that *ab = e*. The remaining element *c* must therefore be its own inverse - *a* and *b* are already taken, after all. This means *c^2 = e* That's actually all that can happen, either all elements are self inverse, or one pair of elements are happily married with the other left to his own devices, pardon the depressing analogy. You might be thinking: 'What if *a* was the self inverse element instead?' This brings me back to the earlier point - the specific names aren't that relevant, what matters is the structure i.e. how they relate to one another. Or you could take the point from the video - any 2 groups with the same Cayley table are 'isomorphic', which essentially means they're the 'same', structurally at least. Now, what can these groups represent? Whenever you have groups of some finite order *n*, you can be assured that the integers mod *n* is always a valid group (or Z/nZ if you want the symbols). This is easy to check, and I'll leave it to you to confirm that the group axioms (closure, identity, associativity and inverses) actually hold. In this case, the group where *ab = c^2 = e* is isomorphic to the integers mod 4, with *c* being the number 2, as double 2 is 0, the identity mod 4. (it's also isomorphic to the group of 4 complex units - namely 1, -1, i, -i under multiplication, with -1 being the self inverse element) The best isomorphism I have for the other group is 180 degree rotations in 3D space about 3 orthogonal axes (say *x*,*y* and *z*). Obviously each element here is self-inverse, as 2 180 degree rotations make a 360 degree rotation, which is the identity. It's easy to check that combining any 2 gives you the other, so the group is closed. I wasn't able to come up with any others, though I'm sure there's a nicer one. As for 5 elements? I only found 2, one of which was non-abelian. One had all elements as self-inverse, the other had 2 pairs of elements that were inverses of each other. The latter is isomorphic to Z/5Z but I've got no idea what the other is isomorphic to. Never mind, the other one isn't even a group - you need to check associativity to be safe. It's a valid operation table, but not for a group unfortunately. It does happen to be a *loop*, which essentially means a group, but less strict, in that associativity isn't necessary. There's an entire 'cube' of different algebraic structures with a binary operation, it turns out, going from the simplest being a magma, to the strictest being a group (and I suppose abelian groups are even stricter). By cube I mean that each structure is positioned at a vertex, with arrows indicating what feature is being added e.g. associativity, identity etc. Wow that was a lot.
@stirlingblackwood
Жыл бұрын
Do you know where I can find a picture of this cube?? Sounds both fascinating and like it would give some interesting context to groups.
@fahrenheit2101
Жыл бұрын
@@stirlingblackwood The wiki article for "Abstract Algebra" has the cube if you scroll down to "Basic Concepts" It's been a while since I looked at this stuff though haha - I'm finding myself reading my own comment and being intimidated by it...
@stirlingblackwood
Жыл бұрын
@@fahrenheit2101 Oh boy, now you got me down a rabbit hole about unital magmas, quasigroups, semigroups, loops, monoids...I need to go to bed 😂
@RISHABHSHARMA-oe4xc
5 ай бұрын
@@fahrenheit2101 bro, are you a Math major ?
@fahrenheit2101
5 ай бұрын
@@RISHABHSHARMA-oe4xc haha, I am now, but wasn't at the time. at the time, I think I was just about to start my first term. I know a fair bit more now, for example, any group of prime order must be cyclic. That said, I do need to brush up on Groups, been a while since I looked at it.
@vanguard7674
8 жыл бұрын
Thank God Abstract Algebra is back :'''D
@deepakmecheri4668
4 жыл бұрын
May God bless you and your channel with good fortune
@RajeshVerma-pb6yo
4 жыл бұрын
Your Explaination is great... First time I able to understand abstract algebra.... Thank you much.. Infinite good wishes for you...😊
@TheZaratustra12
2 ай бұрын
long live the channel and its charming mathematician! Perfect presentation of the topic! I'm getting surer and surer that I can have the level in Math I want to have.
@ashwini8008
4 ай бұрын
thank you, no words dear teacher, you gave me the confidence to learn math....
@johnmorales4328
6 жыл бұрын
I believe the answer to the challenge question are the groups Z/2Z x Z/2Z and Z/4Z.
@larshizzleramnizzle3748
5 жыл бұрын
Thank you! I would've never thought of that Cartesian product!!
@pbondin
6 жыл бұрын
I think the 4 groups are: 1) e a b c 2) e a b c 3) e a b c 4) e a b c a e c b a b c e a c e b a e c b b c e a b c e a b e c a b c a e c b a e c e a b c b a e c b e a However I can't figure out which 3 are identical
@samoneill6222
6 жыл бұрын
The following PDF will give an explanation as to why 3 of the tables are the same. www.math.ucsd.edu/~jwavrik/g32/103_Tables.pdf The trick is to rename the variables a->b, b->c and c->a, thus creating a new table and then rearrange the rows and columns. For example take table 2 and rename a->b, b->c and c->a which generates: e b c a b c a e c a e b a e b c Reorder the rows: e b c a a e b c b c a e c a e b Reorder the columns: e a b c a c e b b e c a c b a e Which is the same as table 3. Effectively the table is disguised by different names for the elements. You can repeat the process with a different naming scheme to see the tables 2,3,4 are all identical. If you try the same trick to table 1 (identity on the diagonal) you will find you just end up with table 1 again. Hence the 2 distinct tables.
@rikkertkoppes
5 жыл бұрын
Note that there is only one with 4 e's on the diagonal. Think about what that means
@hemanthkumartirupati
5 жыл бұрын
@@samoneill6222 Thanks a lot for the explanation :)
@hemanthkumartirupati
5 жыл бұрын
@@rikkertkoppes I am not able discern what that means. Can you help?
@fishgerms
5 жыл бұрын
@@hemanthkumartirupati In the one with e's on the diagonal, each symbol is its own inverse. A * A = E, B * B = E, and C * C = E. In the other groups, there are two symbols that are inverses of each other, and one that's its own inverse. In group 2), A * C = E, and B * B = E. For the other groups, there are also 2 symbols that are inverses of each other, and one that's its own inverse. So, they're the same in that you can swap symbols around and get the same group. For example, group 3) has A * B = E and C * C = E. If you swap symbols B and C, you get A * C = E and B * B = E, which are the same as group 2).
@pinklady7184
3 жыл бұрын
I am learning fast with you. Thank you for tutorials,
@chrissidiras
5 жыл бұрын
Oh dear god, this is the first time I actually engage to a challenge offered in a youtube video!
@andrewolesen8773
6 жыл бұрын
I did the excercise found the groups by setting, a^-1=b, a^-1=c, b^-1=c and finally for the trivial group a^-1=a and b^-1=b and c^-1=c. Came up with four unique Cayley tables though. Don't have 3 equal to each other, wondering where I went wrong.
@stefydivenuto3253
2 жыл бұрын
also I have the same result....3 different group....also I wondering where I went wrong....someone can help me?
@PunmasterSTP
3 жыл бұрын
Those "contradiction" sound effects... But on a more serious note, it took me *so* long to piece these things together on my own. I *really* wish I had found Socratica years ago!
@thegenerationhope5697
6 ай бұрын
What a crystal clear explanation. Really enjoyed the explanation here.
@hectornonayurbusiness2631
4 жыл бұрын
I like how these videos are short. Helps it be digestible.
@eshanene4598
4 жыл бұрын
Excellent video. Way better than most college professors. I think, these videos should be named as "demystifying abstract algebra" or rather "de-terrifying abstract algebra"
@mingyuesun3214
6 жыл бұрын
the background music makes me feel quite intense and wakes me up a lot hahhah. thnak you
@reidchave7192
4 жыл бұрын
That sound when the contradiction appears after 2:50 is hilariously serious
@danielstephenson146
3 жыл бұрын
@ortomy I was looking for someone to comment this hah scared me too!
@hashirraza6461
6 жыл бұрын
You teached in such a fantastic way that it is whole conceptualized.... And in the classroom the same topic is out of understanding! Love u for having such scientific approch...! ❤
@Zeeshan_Ali_Soomro
4 жыл бұрын
The background music in the first part of video plus the way in which socratica was talking was hypnotizing
@paulcohen6727
Күн бұрын
How can a group have an order of less than three and still be associative? I wasn't sure, so I checked and found that the "Associative law states that when three real numbers are added or multiplied together, then the grouping of the numbers does not matter." This implies three different elements unless I'm misreading it. Could {e * e} * { e} = {e} * {e * e} be a valid expression of associativity?
@NaimatWazir0347
5 жыл бұрын
style of your teaching and delivery of lecture are outstanding Madam Socratica
@Socratica
2 жыл бұрын
Socratica Friends, we're excited to share our FIRST BOOK with you! How To Be a Great Student ebook: amzn.to/2Lh3XSP Paperback: amzn.to/3t5jeH3 or read for free when you sign up for Kindle Unlimited: amzn.to/3atr8TJ
@amrita3272
3 ай бұрын
I am watching this in 2024 and it's very helpful.Thank you very much
@aibdraco01
5 жыл бұрын
Thanks a lot for a clear explanation although the topic is so confusing and hard. God bless you !!!
@readjordan2257
Жыл бұрын
Thanks, i just had this review on the midterm about it today and now its in my reccomend. Very apt.
@markmajkowski9545
5 жыл бұрын
Thanks Soln pretty easy GOOD clue The three identical solns take your 3 element group eAB add C*C must be e CB is A and CA B. Then exchange A for C then B for C. That’s 3 which are the same except ordering. Then for the non identical AA=BB=CC=e AC=B AB=C BC=A. This might seem like you can make 3 of these but you cannot. As the first non identity element times the second must be the third, etc so you get only one soon as ordered. In the first you get the identity element as AA BB then CC but these are the same. Fun!
@JamesSpiller314159
4 жыл бұрын
Excellent video. Clear, effortless, and instructive.
@ibrahimn628
4 жыл бұрын
She should be awarded for the way she explained this concept
@randomdude9135
5 жыл бұрын
Thank you. This was an eye opener thought provoking video which cleared many of my doubts which I was searching for.
@sangeethamanickam6002
2 жыл бұрын
U
@jeremylaughery2555
3 жыл бұрын
This is a great video that demonstrates the road map to the solution of the RSA problem.
@1DR31N
4 жыл бұрын
Wished I had you as my teacher when I was at school.
@laurakysercallis653
4 жыл бұрын
Hi! I'm trying to brush up and I have a question about inverses. In a finite group, it seems intuitive that the right and the left inverses probably have to be the same, otherwise there would be probably redundancies, right? I also remember from linear algebra that there is some proof that if right and left inverses both exist for matrices, then the left and right inverses are equal to each other, but I remember that proof felt specific to matrices. If you've got a non commutative infinite group, why is it that the left and right inverses have to be equal to each other? For that matter, is it possible that you could have a right identity and a different left identity? It seems like identity is just "do nothing," but when I got to quotient groups and cosets, the meaning of identity is no longer "do nothing."
@MuffinsAPlenty
4 жыл бұрын
This is a great question! I'll start with the identity question. You cannot have different left- and right-identities. If you have both a left-identity and a right-identity, then they must be equal to each other. To show why, let e be a left-identity and ε be a right-identity. The key is to look at the product eε. Since e is a left-identity, eε = ε. On the other hand, since ε is a right-identity, eε = e. So we have e = eε = ε. So it is impossible to have different left- and right-identities. The "identity" property does not allow it. For the question about inverses, uniqueness is guaranteed by the _associative_ property. So, for example, let's say you have a multiplicative identity element e. Suppose you have an element g which has left inverse h and right inverse k. Then the proof that h = k follows from looking at hgk in two different ways: h(gk) = he = h (hg)k = ek = k So since multiplication is associative, you get h = h(gk) = (hg)k = k. There are group-like structures where we don't require the associative property (these are known as loops and quasigroups). In these structures, it is entirely possible to have an element whose left- and right-inverses are different! Even if you have a finite loop, for example, it's possible that the left- and right-inverses are different from each other.
@jonpritzker3314
2 жыл бұрын
You guys r cool
@Redeemed_Daughter
2 жыл бұрын
When checking for groups G of order 2 , I used the the integers 0 and 1 under addition operation and I don't see how adding 1 with 1 equates to 0. I feel compelled to say 2. But then two is not in the group elements. Where am I going wrong about this??
@paulmccaffrey2985
Жыл бұрын
I'm glad that Arthur Cayley was able to speak at the end.
@xreed8
4 жыл бұрын
The Cayley table for {1,-1,i, -I} is wrong? The multiplication of (1 x i) like you said is that element, so its 1, not i, in the table. Furthermore, how is (i x i) = -1? What is i?
@MuffinsAPlenty
4 жыл бұрын
i is what is often referred to as an "imaginary" or "complex" number. i is a number with the property that i^2 = -1. In this context, i does not stand for "identity" but rather a number which squares to -1. 1 is the identity under multiplication.
@AMIRMATHs
2 жыл бұрын
Thenks so much ...im following you from Algeria 🇩🇿
@Socratica
2 жыл бұрын
Hello to our Socratica Friends in Algeria!! 💜🦉
@antoniusnies-komponistpian2172
10 ай бұрын
The one group of order 4 is addition in Z/4Z, the other one is the standard base of the quaternions without signs
@نظورينظوري-ز2ظ
5 жыл бұрын
راءع جدا افتهموت اكثر من محاضرات الجامعة لان بالمحاضرة انام من ورة الاستاذ ساعة يلا نفتهم منة معنى الحلقة
@fredm73
11 ай бұрын
A question: must every valid Cayley table (as defined in video) represent a valid group? I.E. does associative law necessarily hold?
@Manuelsanchez-eu7ez
10 ай бұрын
All groups are associative by definition
@mayurgare
3 жыл бұрын
The explanation was so simple and easy to understand. Thank You !!!
@pasanrodrigo3463
3 жыл бұрын
No chance of getting an unsubscribed fan !!! 1.Veeeeeeery Clever 2.Ending of the video Booms!!!
@subramaniannk4255
8 ай бұрын
The best video on Cayley Table..it got me thinking
@ABC-jq7ve
Жыл бұрын
Love the vids! I’m binge watching the playlist before the algebra class next semester :D
@fvanessagan
3 жыл бұрын
Ok so did the identity in a Cayley Table always exist in each rows and columns? Because if you take the multiplication table (integer). There are no 1 in the second row. Or does it apply only for some?
@aweebthatlovesmath4220
2 жыл бұрын
This video was so beautiful that i cannot describe it with words.
@twostarunique7703
5 жыл бұрын
Excellent teaching style
@annievmathew5361
3 жыл бұрын
Pls include a video on how to find the generators of a cyclic group of multiplicative order
@mksarav75
6 жыл бұрын
What a beautiful way to teach abstract algebra! Thanks a lot.
@izzamahfudhiaaz-zahro7949
Жыл бұрын
hallo, i'm from indonesia and i like your videos, thanks you
@MUHAMMADSALEEM-hu9hk
5 жыл бұрын
thanks mam .your lecture is very helpful for me
@b43xoit
5 жыл бұрын
Is there a shortcut for checking a table for associativity? Or is it necessary to try every possible combination of three choices of element?
@MuffinsAPlenty
5 жыл бұрын
No, it's a hard problem! There are some facts you can use to slightly reduce the number of things you have to check. For example, for any element x, it is automatically true that x*(x*x) = (x*x)*x, so you don't have to check triples of the same element. Additionally, if you can identify that you have a two-sided identity element in your table, you can't have to check _anything_ with the identity element. Say e is the identity element. Then, for example, e*x = x, so (e*x)*y = x*y = e*(x*y) And you can change the position of e here. There are also some warning signs that can make it obvious that a multiplication table is _not_ associative. For example, if you know you have an associative property, you can prove that inverses are two-sided and unique. But for non-associative operations, the left inverse of an element _can_ be different from the right inverse of the same element. So if you have an identity element e so that xy = e and zx = e with y ≠ z, then you know right away that the multiplication is _not_ associative. Unfortunately, this test is not an "if and only if" test. It is possible that all inverse are unique with multiplication _still_ not associative.
@MuffinsAPlenty
4 жыл бұрын
Hello! I wanted to leave another comment pointing out a few more things. When I said you don't have to check x*(x*x) = (x*x)*x, I was wrong about that. In general, you _do_ need to check that. The scenario in which you don't need to check it is if you know that your operation is commutative. It's not hard to check commutativity on a Cayley table - just see if the table is symmetric about its diagonal. Also, it was recently brought to my attention that there is a more structured way for checking associativity, called Light's associativity test. For more information about Light's associativity test, check out the Wikipedia article. en *[dot]* wikipedia *[dot]* org/wiki/Light%27s_associativity_test
@grexxiogdgd376
4 жыл бұрын
So operation tables are just abstract sudokus
@RedefiningtheConcepts
6 жыл бұрын
It was very very good so never stop.
@utkarshraj4268
7 ай бұрын
This is really helpful Love from india 🇮🇳🇮🇳
@cameronramsay118
5 жыл бұрын
This was a very abstract excel tutorial
@rayrocher6887
7 жыл бұрын
this was helpful as a keystone to abstract algebra, thanks for the encouragement.
@minhazulislam4682
2 жыл бұрын
so, I used a pro gamer move to find the caley table of order 4. I basically created Z mod 4 table and changed 0,1,2,3 to e,a,b,c respectively. It worked!
@iyaszawde
2 жыл бұрын
Thanks for all vedios you made, they are so exciting and easy to understand ❤❤
@jadeconjusta1449
3 жыл бұрын
i love the sound fx everytime there's a contradiction
@pittdancer85
Жыл бұрын
Would it be correct to say the 4 groups would be a^2=e, b^2=e, c^2=e, and a^2=b^2=c^2=e? The first three are actually identical.
@satyamrai39
5 жыл бұрын
So. Firstly I constructed the outer most (left corner ) of the table... Then just thought because of the pattern that z mod 4 table will be a valid one which it was! (Of course )... Then I realized that the inner 3x3 table of the zmod 4 table(one formed by excluding the top row and column)can be rotated to get 4 different new tables ..(flipping just gives same thing and rotation followed by flip just gave one of those 4 tables).. Then. Only two of them... contained all the elements once... Which was the solution...(which is already mentioned by others ) Woah.. just amazing vid.. (Hmm so what about 5x5?)
@satyamrai39
5 жыл бұрын
Gosh just found out that this vid was 3years ago😂
@Socratica
5 жыл бұрын
We're so happy you've found us!! That's the nice thing about making videos about math & science - every year we have a chance to help new students.
@sarahdanielleanderson9772
Жыл бұрын
I need help... I created four different Cayley tables for a group with order four. There are no duplicate elements in any row or column. I am wondering how we know three of them are the same? I looked through them and even thought about switching rows or columns around, and I don't see it.
@MuffinsAPlenty
Жыл бұрын
One super tricky thing about trying to construct groups by building Cayley tables is checking the associative property. There aren't many easy checks for the associative property when looking at a Cayley table. Luckily, with four elements, it isn't _too_ bad to check whether your operation is associative or not. I recommend checking each of your tables for the associative property. Check whether x(yz) = (xy)z for all x, y, and z you have (this will include repeat elements!). In total, this would be 64 checks since there are four elements, so you have 4 choices for x, 4 choices for y, and 4 choices for z. But you may be able to narrow down how many checks you do. For example, if your table has an identity element, then you don't have to check any computation involving the identity element. For example, e(xy) = xy = (ex)y follows from the property of e being the identity. And you can, similarly, show that (xe)y = x(ey) and (xy)e = x(ye) also hold for free, regardless of what x and y are. So if you eliminate the identity element, you only have three choices for x, y, z, giving us 27 checks in total. Another possible way to reduce the number of checks you have to do is if you can see that your table is commutative, then you don't have to check "cubes" - that is, you don't have to check x(xx) = (xx)x for some fixed element x, because knowing the multiplication is commutative gives us x(xx) = (xx)x for free. So if your table has an identity (and it should!), then that reduces the checks down to 27, and if it's also commutative, than that reduces the checks down 3 more to 24. Unfortunately, there's not much more you can do to reduce number of checks. Feel free to comment back if you have any further questions!
@humamalsebai
7 жыл бұрын
It is worth mentioning that the fact that a group contains no duplicate elements in any row or column is referred to as the "latin square" property. It is also important to realize, for a group that satisfies the associativity property, the inverse property and the :identity element property then that group is a latin square. This is evident in the video at 2:41 where all of the previously mentioned property are invoked in proving the latin square property. However, there are some latin square (quasigroups) that are not groups. Not every magma that satisfies the latin square property is a group. In this case the quasigroup is said to have the invertibility property ( not the inverse property)
@jonpritzker3314
2 жыл бұрын
What does molten rock not exposed to open air have to do with this?
@balthazarbeutelwolf9097
4 жыл бұрын
when "searching" for groups by generating these tables do we need check that associativity holds, or do we get this for free? If we do not get it for free (which is what I would suspect) do we get for free that left and right inverses are the same? Suppose that was not true either, and we would specifically check our tables for that property - would that be sufficient to achieve associativity?
@MuffinsAPlenty
4 жыл бұрын
Checking tables for associativity is _very_ hard. In _general,_ you really do need to check every triple of elements to check for associativity, since it's possible to define a binary operation on a set with at least 3 elements which is associative for all triples of elements _except one._ Your left and right inverses comment is actually useful. Associativity requires left and right inverses to be the same. So if you notice that an element has a different left inverse from its right inverse, you automatically know that the operation is _not_ associative. The backwards implication is _not_ true, however. It's entirely possible for a non-associative operation to have an identity and for every element to have a unique two-sided inverse. There is an algorithm to sort of "streamline" checking for associativity. It is known as Light's Associativity Test. It's more or less and organized way to compare all triples.
@ZVENOM6134
3 жыл бұрын
Hmmm I don't quite understand why a*a is e for the order of 2. I understand that you're using the "rules" made from previous observations but to contextualise them into lets say e being 1 and a being 2, it wouldn't add up :/ Or am I seeing this in the wrong way?
@cindarthomas3584
3 жыл бұрын
Thank you soo much 💝💝 I'm not able to express my gratitude.. your videos made me love algebra.. Earlier I didn't like it
@waynelast1685
4 жыл бұрын
these videos very well written so far
@julianocamargo6674
2 жыл бұрын
Best explanation in the world
@alejrandom6592
2 ай бұрын
The two groups are {1,i,-1,-i} and {1,j,-1,-j} from split complex numbers
@alejrandom6592
2 ай бұрын
First is rotations of a square, second is symmetries of a rectangle (allowing flipping)
@Nekuzir
2 жыл бұрын
Curiosity has me learning about octionions and above, this video is helpful in that endeavor
@prodipmukherjee2218
6 жыл бұрын
It's very helpful for everyone interested in mathematics.
@waynelast1685
4 жыл бұрын
So not mentioned so far I think is : elements in a group do NOT necessarily have to be sequential. For example, for group order 3 integers under addition, you can have elements 0,2,4 or 0,100, 150 ?
@Edgawliet
2 жыл бұрын
I get with this technique you could build a set and a relation that is 1) close 2) with a unique neutro element 3) with unique inverses, but how you know the tables made this way always satisfy the associativity? hence, they represent a group.
@sukannyanath5583
5 жыл бұрын
Intergers modulo 4 under addition and klein 4 group will be the answer
@saharupam29
6 жыл бұрын
e a b c e e a b c a a e c b b b c e a c c b a e Soothing lectures.. Really had a fun with these abstract things
@bunnnieee7163
2 жыл бұрын
Hi I'm confused in making cayley tables for set {-1,0,1} is it possible when using addition as a binary operation? Thank youu
@DeekshaVerma-i2s
Жыл бұрын
{1,0,-1} is not a group under either multiplication or addition.
@emmadurza3
4 жыл бұрын
Shouldn't the unique order 2 group be isomorphic to the integers mod 2 *under addition*? Shouldn't we mention an operation if it's a group?
@MuffinsAPlenty
4 жыл бұрын
Sure, it might make things clearer to mention the operation. But there's really only one "natural" operation on Z/2Z that makes it into a *group,* and that' addition. Z/2Z is *not* a group under multiplication.
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