Group Multiplication Tables | Cayley Tables (Abstract Algebra)

[@Socratica]

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[@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.

[@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.

[@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.

[@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

[@kingston9582]

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

[@kirstens1389]

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

[@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!

[@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 ^^

[@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.

[@fg_arnold]

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

[@sandeepk4339]

I'm from India, your explanation was outstanding.

[@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!

[@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

[@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!

[@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.

[@arrpit5774]

Just loved your content , getting easier with each passing minute

[@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...

[@JJ_TheGreat]

This reminds me of Sudoku! :-)

[@eshanene4598]

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"

[@deepakmecheri4668]

May God bless you and your channel with good fortune

[@thegenerationhope5697]

What a crystal clear explanation. Really enjoyed the explanation here.

[@RajeshVerma-pb6yo]

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

[@vanguard7674]

Thank God Abstract Algebra is back :'''D

[@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.

[@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!

[@hectornonayurbusiness2631]

I like how these videos are short. Helps it be digestible.

[@chrissidiras]

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

[@Zeeshan_Ali_Soomro]

The background music in the first part of video plus the way in which socratica was talking was hypnotizing

[@pasanrodrigo3463]

No chance of getting an unsubscribed fan !!! 1.Veeeeeeery Clever 2.Ending of the video Booms!!!

[@ibrahimn628]

She should be awarded for the way she explained this concept

[@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!

[@paulmccaffrey2985]

I'm glad that Arthur Cayley was able to speak at the end.

[@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.

[@readjordan2257]

Thanks, i just had this review on the midterm about it today and now its in my reccomend. Very apt.

[@ashwini8008]

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

[@NaimatWazir0347]

style of your teaching and delivery of lecture are outstanding Madam Socratica

[@jeremylaughery2555]

This is a great video that demonstrates the road map to the solution of the RSA problem.

[@pinklady7184]

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

[@1DR31N]

Wished I had you as my teacher when I was at school.

[@JamesSpiller314159]

Excellent video. Clear, effortless, and instructive.

[@hashirraza6461]

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...! ❤

[@subramaniannk4255]

The best video on Cayley Table..it got me thinking

[@AMIRMATHs]

Thenks so much ...im following you from Algeria 🇩🇿

[@Socratica]

Hello to our Socratica Friends in Algeria!! 💜🦉

[@aweebthatlovesmath4220]

This video was so beautiful that i cannot describe it with words.

[@ABC-jq7ve]

Love the vids! I’m binge watching the playlist before the algebra class next semester :D

[@aibdraco01]

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

[@adhithyalaxman4094]

This channel is just wayy too good! :)

[@mayurgare]

The explanation was so simple and easy to understand. Thank You !!!

[@Socratica]

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[@julianocamargo6674]

Best explanation in the world

[@randomdude9135]

Thank you. This was an eye opener thought provoking video which cleared many of my doubts which I was searching for.

[@sangeethamanickam6002]

U

[@waynelast1685]

these videos very well written so far

[@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

[@RedefiningtheConcepts]

It was very very good so never stop.

[@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!

[@cindarthomas3584]

Thank you soo much 💝💝 I'm not able to express my gratitude.. your videos made me love algebra.. Earlier I didn't like it

[@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??

[@twostarunique7703]

Excellent teaching style

[@mksarav75]

What a beautiful way to teach abstract algebra! Thanks a lot.

[@jadeconjusta1449]

i love the sound fx everytime there's a contradiction

[@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.

[@rayrocher6887]

this was helpful as a keystone to abstract algebra, thanks for the encouragement.

[@iyaszawde]

Thanks for all vedios you made, they are so exciting and easy to understand ❤❤

[@AnuragSingh-ds7db]

Big fan of you... you explained very well❤❤

[@utkarshraj4268]

This is really helpful Love from india 🇮🇳🇮🇳

[@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?

[@arunray5365]

You teaching style is awesome

[@izzamahfudhiaaz-zahro7949]

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

[@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!!

[@owlblocksdavid4955]

I watched some of these for fun before. Now, I'm coming back to supplement the set theory in my discrete mathematics textbook.

[@Nekuzir]

Curiosity has me learning about octionions and above, this video is helpful in that endeavor

[@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?

[@prodipmukherjee2218]

It's very helpful for everyone interested in mathematics.

[@drsamehelhadidi9609]

Very nice explanation

[@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!

[@zubairjaved3123]

So beautiful explanation

[@نظورينظوري-ز2ظ]

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

[@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

[@vs6x3]

5:59 symmetric (along the diag)how do you mean?🤔😐

[@ertwro]

If you look at the elements is like there's a diagonal mirror. So if you number the elements: 1 2 3 e a b 4 5 6 => a b e 7 8 9 b e a 4=2, the a 3=7, the b 6=8, the e of course, 1,5,9 are themselves so their hypothetical halves are symmetrical.

[@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

[@universeandparticles]

Legend in mathematics😍😍

[@b43xoit]

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]

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]

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

[@narendrakhadka9598]

Excellent.i learned very clearly algebra.

[@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

[@annievmathew5361]

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

[@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).

[@mosca-tse-tse]

Loved it. So beautifully explained. 👌

[@MUHAMMADSALEEM-hu9hk]

thanks mam .your lecture is very helpful for me

[@AdolfNdlovu]

Thank you for this video. It is really helpful

[@divyadulmini374]

Thank you very much.I understood the lesson easily ❤️❤️❤️

[@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.)

[@sarahdanielleanderson9772]

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]

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!

[@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!

[@missghani8646]

you are fun to watch, really you are doing a great job, abstract algebra was never fun. Thank you

[@MondoDuplantis354]

We need your classes ❤

[@jumpoutatree]

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

[@iconicarts6268]

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