Group Multiplication Tables | Cayley Tables (Abstract Algebra)

[@Socratica]

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[@kirstens1389]

These videos are really extremely helpful - too good to be true - for learning overall concepts.

[@sadiqurrahman2]

You explained a confusing topic in the most easiest manner. Thanks a lot.

[@zy9662]

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]

@@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]

@@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]

@@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]

@@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.

[@fredm73]

A question: must every valid Cayley table (as defined in video) represent a valid group? I.E. does associative law necessarily hold?

[@Manuelsanchez-eu7ez]

All groups are associative by definition

[@filipve73]

(abstract) is there a relationship between "group multiplication tables" and "truth tables" ?? (end abstract)

[@LawsonEnglish]

Don't think so as the rules to construct the table are completely different. You could define a truth table to look like a group multiplication table in some cases, though and maybe there are special kinds of situations where that is important.

[@b43xoit]

My quick thought, not carefully checked, says exclusive-or is a group, but AND and OR are not. In fact, the video says there is only one group of order 2. I think you will find it is exactly XOR. e=FALSE; a=TRUE.

[@vs6x3]

🤔1:51 you are you talking about diag elements, I took Group ×{1, -1, 2, -2} why is it not giving me Identity Element?😢

[@nagys36snn]

How do you get to 2

[@yvanbrunel9734]

the weird thing is I have to convince myself that "+" doesn't mean "plus" anymore 😩

[@Abhishek._bombay]

Addition modulo 🙌😂

[@jason-mr]

@@yvanbrunel9734 what do you mean?

[@mehulkumar3469]

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.

[@JJ_TheGreat]

This reminds me of Sudoku! :-)

[@MoayyadYaghi]

I literally went from Struggling in my abstract algebra course to actually loving it !! All love and support from Jordan.

[@Socratica]

This is so wonderful to hear - thank you for writing and letting us know! It really inspires us to keep going!! 💜🦉

[@tristanreid]

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]

She says that there are 2 distinct groups because 1 is abelian and the rest of them are normal groups.

[@rajeevgodse2896]

@@dunisanisambo9946 Actually, all of the groups are abelian! The smallest non-abelian group is the dihedral group of order 6.

[@jonpritzker3314]

Your comment helped me without spoiling the fun :)

[@fahrenheit2101]

@@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]

@@rajeevgodse2896 Nevermind, turns out I needed to check associativity - I'm surprised that isn't a given.

[@waynelast1685]

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]

She says actually a times e, but here order it's important. And yes you are allright.

[@kingston9582]

This lesson saved my life omg. Thank you so much for being thorough with this stuff, my professor was so vague!

[@fahrenheit2101]

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]

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]

@@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]

@@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]

@@fahrenheit2101 bro, are you a Math major ?

[@fahrenheit2101]

@@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.

[@grexxiogdgd376]

So operation tables are just abstract sudokus

[@efeuzel1399]

I am watching and liking this in 2020!

[@markpetersenycong8723]

Guess we are here because of online class due to the Covid-19 😂

[@halilibrahimcetin9448]

Been to math village in Turkey?

[@sukhavaho]

@@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]

i am in2023

[@АялаБақытбек]

2023...

[@mheermance]

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.

[@fg_arnold]

love the Gilliam / Python allusions at the end. good work Harrisons, as usual.

[@sandeepk4339]

I'm from India, your explanation was outstanding.

[@SaebaRyo21]

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 ^^

[@amrita3272]

I am watching this in 2024 and it's very helpful.Thank you very much

[@hansteam]

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!

[@vanguard7674]

Thank God Abstract Algebra is back :'''D

[@pbondin]

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]

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]

Note that there is only one with 4 e's on the diagonal. Think about what that means

[@hemanthkumartirupati]

@@samoneill6222 Thanks a lot for the explanation :)

[@hemanthkumartirupati]

@@rikkertkoppes I am not able discern what that means. Can you help?

[@fishgerms]

@@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).

[@TheFhdude]

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]

But how do you know that the associative law holds?

[@jonatangarcia8564]

@@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]

@@jonatangarcia8564 Yeah how do you prove that the cayley table made by following the rules said by her always follows the associative law?

[@jonatangarcia8564]

@@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

[@youtwothirtyfive]

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!

[@JozuaSijsling]

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.

[@johnmorales4328]

I believe the answer to the challenge question are the groups Z/2Z x Z/2Z and Z/4Z.

[@larshizzleramnizzle3748]

Thank you! I would've never thought of that Cartesian product!!

[@xreed8]

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]

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.

[@andrewolesen8773]

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]

also I have the same result....3 different group....also I wondering where I went wrong....someone can help me?

[@Redeemed_Daughter]

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??

[@reidchave7192]

That sound when the contradiction appears after 2:50 is hilariously serious

[@danielstephenson146]

@ortomy I was looking for someone to comment this hah scared me too!

[@tomasito_2021]

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

[@mingyuesun3214]

the background music makes me feel quite intense and wakes me up a lot hahhah. thnak you

[@PunmasterSTP]

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!

[@arrpit5774]

Just loved your content , getting easier with each passing minute

[@amyathas.pradeep6949]

Omg... what the hell was i studied in my college 😂😂😂

[@RajeshVerma-pb6yo]

Your Explaination is great... First time I able to understand abstract algebra.... Thank you much.. Infinite good wishes for you...😊

[@ozzyfromspace]

SPOILER ALERT: ANSWER. (Click read more for answer) There are two groups of order 4. Any other groups are isomorphic wrt. these two groups. let e = 0, a = 1, b = 2, c = 3: Solution 1: 0 1 2 3 1 0 3 2 2 3 1 0 3 2 0 1 Solution 2: 0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0 You can generate these Cayley tables by trial and error, doing tree diagrams. Whenever you generate multiple solution from the same branch, the solutions are independent (think about it, it should make sense). Whenever you generate solutions from other branches, the solution set may contain isomorphisms. To check for isomorphisms, performs substitutions a b on a random group and reorganize the rows AND columns so your elements are ordered as before. The operation of a b followed by appropriate reorganizing of rows and columns will generate an isomorphic group. Any groups in the same branch as your isomorphic group can be disregarded. Rinse and repeat for a c. In practice, this is enough to get the answer, but a computer would also check b c. Doing this will leave *Solution 2* and any other group you found. *Solution 1* is just a 'natural' choice because it was generated from the same logical branch as *Solution 2.* I hope you found this helpful. Excellent presentation, Socratica! I used to generate puzzles like this when I was a kid, and did not realize they were called Cayley Multiplication Tables. You brought back fun memories. I even have a theorem about them in terms of math I created for myself from years back (and equation I haven't been able to solve). If you're interested, check out this open problem I posted to Stack-Exchange a year or two ago: math.stackexchange.com/q/2845247 Liliana de Castro and Team, you're awesome! -Float.

[@KianAzami]

if we consider the group elements: e, a, b, c. Then: 1. { e, a, b, c a, e, c, b b, c, e, a c, b, a, e } 2. { e, a, b, c a, b, c, e b, c, e, a c, e, a, b } 3. { e, a, b, c a, c, e, b b, e, c, a c, b, a, e } 4. { e, a, b, c a, e, c, b b, c, a, e c, b, e, a } These are the 4 groups. When we can say that two groups are identical?

[@urieldavid]

I have the same answer and then I have the same question.

[@Klarpimier]

3:55 EA SPORTS It’s in the game

[@slingshotninja6970]

watch the first 2 mins and skip to 4;00 You'll understand..

[@laurakysercallis653]

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]

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]

You guys r cool

[@A7medKhairy]

drive.google.com/open?id=0B4MhvcWVT67RU1FkSTJMVWY2MHM ^^

[@oximas]

great job that's the same as what i did

[@urieldavid]

I can make this too but I don't know what is cyclic. Why are there only 2 distinct groups?

[@MoonLight11023]

The simplest way to solve the puzzle is to use the group (Z4 , +) and then substitute alphabets for numbers. Z4 + 0 1 2 3 * e a b c This is called cheating guys. Actually it is an abelian group. 0 0 1 2 3 e e a b c 1 1 2 3 0 a a b c e 2 2 3 0 1 b b c e a 3 3 0 1 2 c c e a b The second one is called Klein four-group; it is abelian but NOT cyclic. + e a b c e e a b c a a e c b b b c e a c c b a e

[@نظورينظوري-ز2ظ]

راءع جدا افتهموت اكثر من محاضرات الجامعة لان بالمحاضرة انام من ورة الاستاذ ساعة يلا نفتهم منة معنى الحلقة

[@ivoriankoua3916]

Who sub due to Cayley ?

[@randomdude9135]

I'd subbed earlier

[@markmajkowski9545]

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!

[@deepakmecheri4668]

May God bless you and your channel with good fortune

[@waynelast1685]

OMG it is just like soduko!

[@MrCEO-jw1vm]

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!

[@kasturiranjanswain9808]

how the three tables are identical

[@chrissidiras]

Oh dear god, this is the first time I actually engage to a challenge offered in a youtube video!

[@pinklady7184]

I am learning fast with you. Thank you for tutorials,

[@keplergelotte7207]

Ni!!

[@satyamrai39]

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]

Gosh just found out that this vid was 3years ago😂

[@Socratica]

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.

[@ozzyfromspace]

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!

[@tnp1022]

or you can say ab is not equal b because a is not the identity element since e is already our identity element.

[@aibdraco01]

Thanks a lot for a clear explanation although the topic is so confusing and hard. God bless you !!!

[@philosophyversuslogic]

I beg your pardon, but is repeating elements in a tablet derives exactly a contradiction? I mean the case: x*a=y*a ... x=y? Because it would be, if previously x=y could be taken as a contradiction. Until we don't know what are these 'x' and 'y', the case 'x=y' is just a certain case. I guess there must be some kind of the restriction before getting such a result: we must define what is a contradiction for a certain tablet, and after that to go further. Again, why I insist on that here: if to assume the Leibniz's law of indescernibles, there are none. According to the one for all x and y [(x=y)->(Fx=Fy)] and here we see the same.

[@adityakhanna113]

So is a sudoku a group of order 9?

[@tresteinjordklatt8133]

No, usually it won't have an identity element.

[@ctogaurav]

Lol

[@ctogaurav]

Great doubt! But, unfortunately sudoku is not a group of order 9.

[@G5349]

No sudoku is not a group because it does not have any of the properties that make a group a group. Sudoku is just a latin square

[@annievmathew5361]

Pls include a video on how to find the generators of a cyclic group of multiplicative order

[@pittdancer85]

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.

[@keyboard_toucher]

If any row or column in a Cayley table were not a permutation of the group elements, that would mean some element doesn't have a (unique) inverse under the group operation, which violates the definition of a group.

[@sukannyanath5583]

Intergers modulo 4 under addition and klein 4 group will be the answer

[@homiramanuj]

I didn’t get the clues, what do you mean by ALL GROUPS of order 2 we always see single General Group 4 Cayley tables of order 4 Sorry Mam I didn’t get it😅 please explain Anyone!?

[@minhazulislam4682]

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!

[@ZVENOM6134]

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?

[@chair547]

I don't see why you need the Kaylee tables to prove that the integers mod 2 are identical to the group we found. Since we prove that group we found is the only order 2 group and we know z2 is an order 2 group, isn't it trivial?

[@gk-qf9hv]

There is a cercular argument in this video! Sooorry... Namely, you wanted to prove that there is no repetition in any raw or colomn, but then you used the same fact to fill up the boxes.. Maybe this is not what you meant to say??

[@TheZaratustra12]

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.

[@integrantedavidanoturna]

That's pretty cool. Does that mean that every group of order 3 is abelian?

[@iconicarts6268]

How many tables will be there for group of order 5,6,7... etc is there any relation.

[@DeLuini985]

I found the 4 possible Cayley tables, but how do I now determine that 3 are identical?

[@aabidmushtaq3243]

I am watching u r videos in 2020 COVID19 From kashmir

[@EyalRock]

I managed to find the two kind of groups the first kind has 2 elements that inverse each other and 1 that is an inverse of it self and the second kind has 3 elements that are an inverse of them selfs. I tried to find examples of this groups. I found out that integers (mod 4) is the first kind , the group 1,i,-1,-i. also the first kind. I could not find an example of the second kind 🙄 can anyone give me one?

[@aldobernaltvbernal8745]

i found that making a table of XOR with the values e = 00, a = 01, b = 10, c = 11 in binary works as an example of the second kind

[@jumpoutatree]

So for groups of order 4, I obviously did integers mod 4 and got a table, but what were the others??

[@DeekshaVerma-i2s]

I came up with 4 different cayleys table. Can you explain how there would be two tables?

[@julianocamargo6674]

Best explanation in the world

[@ashwini8008]

thank you, no words dear teacher, you gave me the confidence to learn math....

[@mmariokart231]

It’s not obvious to me why the defining difference would end up being what results in e and arbitrary names for those things

[@pulverizedpeanuts]

eabc aceb beca cbae

[@forheuristiclifeksh7836]

2:00 Every row and column contains identity because,

[@pearlairahcinco3868]

Woww thank you so much for thr beautiful explanations

[@tadanohitohito9096]

Am I the only one who heard bloodborne parry sounds on the contradictions ?

[@saharupam29]

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

[@MUHAMMADSALEEM-hu9hk]

thanks mam .your lecture is very helpful for me

[@iconicarts6268]

Can we use this for making different number systems and thire operations?

[@utkarshraj4268]

This is really helpful Love from india 🇮🇳🇮🇳

[@antoniusnies-komponistpian2172]

The one group of order 4 is addition in Z/4Z, the other one is the standard base of the quaternions without signs

[@aniketsaha7455]

Cayley table is for any algebraic structure and not only groups..

[@luckydannumber2]

So Cayley tables are essentially Sudoku.

[@paulcohen6727]

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?

[@MuffinsAPlenty]

Associativity doesn't require distinct elements. For a set with two elements a and b, and a binary operation *, you need to check all of the following: (a*a)*a = a*(a*a) (a*a)*b = a*(a*b) (a*b)*a = a*(b*a) (a*b)*b = a*(b*b) (b*a)*a = b*(a*a) (b*a)*b = b*(a*b) (b*b)*a = b*(b*a) (b*b)*b = b*(b*b) So yes, in a one-element set with "e" as the only element, you would need to check (e*e)*e = e*(e*e). (Of course, if e were an identity, this would be easy to check.)

[@squirrward]

anyone who can suggest a channel that proves statement in abstract algebra?

[@shotacercva4758]

very helpful. THank you so much.

[@ifrazali3052]

4:23 why isn't a squared

[@urieldavid]

because the "product" should be closed in the group - select from the elements only

[@iusyiftgkl7346u]

It is, and a^2 must be an element from the group, what together with the non-repetition properties gives its value

[@izzamahfudhiaaz-zahro7949]

hallo, i'm from indonesia and i like your videos, thanks you

[@b43xoit]

Has group theory been applied to quantum chromodynamics?

[@GregRebholz]

Yes, QCD is non-abelian with symmetry group SU(3)

[@alejrandom6592]

The two groups are {1,i,-1,-i} and {1,j,-1,-j} from split complex numbers

[@alejrandom6592]

First is rotations of a square, second is symmetries of a rectangle (allowing flipping)

[@zeroUnknown117]

Xlelalblcl elelalblcl alalelclbl blblclelal clclblalel This is abelian right? so would e be the identity matrix? e.g. e=1000 0100 0010 0001