@@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
@@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.
@@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.
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).
@@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?
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)
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.
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 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 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
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 ^^
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
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!
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!
@@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!)
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.
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!
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 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...
@@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.
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"
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!
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!
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.
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!
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)
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.
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...! ❤
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
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.
@@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).
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!
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?)
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.
"3 groups are identical" Isomorphic groups. Down the rabbit hole I go. If I weren't for searching for other materials, I would be stuck. Nonetheless, this is an excellent tutorial. "2 distinct groups" One for int mod 4, and one for that K4 group.
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You explained a confusing topic in the most easiest manner. Thanks a lot.
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?
@@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
@@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
@@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.
@@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.
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.
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).
She says that there are 2 distinct groups because 1 is abelian and the rest of them are normal groups.
@@dunisanisambo9946 Actually, all of the groups are abelian! The smallest non-abelian group is the dihedral group of order 6.
Your comment helped me without spoiling the fun :)
@@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?
@@rajeevgodse2896 Nevermind, turns out I needed to check associativity - I'm surprised that isn't a given.
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)
She says actually a times e, but here order it's important. And yes you are allright.
I literally went from Struggling in my abstract algebra course to actually loving it !! All love and support from Jordan.
This is so wonderful to hear - thank you for writing and letting us know! It really inspires us to keep going!! 💜🦉
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.
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!
But how do you know that the associative law holds?
@@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).
@@jonatangarcia8564 Yeah how do you prove that the cayley table made by following the rules said by her always follows the associative law?
@@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
I'm from India, your explanation was outstanding.
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 ^^
These videos are really extremely helpful - too good to be true - for learning overall concepts.
This lesson saved my life omg. Thank you so much for being thorough with this stuff, my professor was so vague!
This reminds me of Sudoku! :-)
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
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!
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!
I am watching and liking this in 2020!
Guess we are here because of online class due to the Covid-19 😂
Been to math village in Turkey?
@@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!)
i am in2023
2023...
love the Gilliam / Python allusions at the end. good work Harrisons, as usual.
No chance of getting an unsubscribed fan !!!
1.Veeeeeeery Clever
2.Ending of the video Booms!!!
Just loved your content , getting easier with each passing minute
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.
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!
What a crystal clear explanation. Really enjoyed the explanation here.
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.
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.
@@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...
@@fahrenheit2101 Oh boy, now you got me down a rabbit hole about unital magmas, quasigroups, semigroups, loops, monoids...I need to go to bed 😂
@@fahrenheit2101 bro, are you a Math major ?
@@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.
May God bless you and your channel with good fortune
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"
Thank God Abstract Algebra is back :'''D
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!
She should be awarded for the way she explained this concept
The background music in the first part of video plus the way in which socratica was talking was hypnotizing
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!
Your Explaination is great...
First time I able to understand abstract algebra....
Thank you much..
Infinite good wishes for you...😊
Thanks, i just had this review on the midterm about it today and now its in my reccomend. Very apt.
I'm glad that Arthur Cayley was able to speak at the end.
thank you, no words dear teacher, you gave me the confidence to learn math....
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.
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!
I like how these videos are short. Helps it be digestible.
Oh dear god, this is the first time I actually engage to a challenge offered in a youtube video!
Thenks so much ...im following you from Algeria 🇩🇿
Hello to our Socratica Friends in Algeria!! 💜🦉
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
2:54 To be more abstract , (a^-1 * a) is e and e*(var) is var
At time mark 6:05, it is better to say one group up to isomorphism (or identical up to isomorphism) rather than identical.
I am learning fast with you. Thank you for tutorials,
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)
What does molten rock not exposed to open air have to do with this?
This is really helpful
Love from india 🇮🇳🇮🇳
Thanks a lot for a clear explanation although the topic is so confusing and hard. God bless you !!!
The best video on Cayley Table..it got me thinking
I believe the answer to the challenge question are the groups Z/2Z x Z/2Z and Z/4Z.
Thank you! I would've never thought of that Cartesian product!!
Wished I had you as my teacher when I was at school.
راءع جدا افتهموت اكثر من محاضرات الجامعة لان بالمحاضرة انام من ورة الاستاذ ساعة يلا نفتهم منة معنى الحلقة
This is a great video that demonstrates the road map to the solution of the RSA problem.
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.
also I have the same result....3 different group....also I wondering where I went wrong....someone can help me?
style of your teaching and delivery of lecture are outstanding Madam Socratica
the background music makes me feel quite intense and wakes me up a lot hahhah. thnak you
This video was so beautiful that i cannot describe it with words.
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...! ❤
hallo, i'm from indonesia and i like your videos, thanks you
Excellent video. Clear, effortless, and instructive.
The one group of order 4 is addition in Z/4Z, the other one is the standard base of the quaternions without signs
Excellent teaching style
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
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.
Note that there is only one with 4 e's on the diagonal. Think about what that means
@@samoneill6222 Thanks a lot for the explanation :)
@@rikkertkoppes I am not able discern what that means. Can you help?
@@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).
these videos very well written so far
Thank you soo much 💝💝
I'm not able to express my gratitude.. your videos made me love algebra..
Earlier I didn't like it
Best explanation in the world
the weird thing is I have to convince myself that "+" doesn't mean "plus" anymore 😩
Addition modulo 🙌😂
@@yvanbrunel9734 what do you mean?
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!
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?)
Gosh just found out that this vid was 3years ago😂
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.
That sound when the contradiction appears after 2:50 is hilariously serious
@ortomy I was looking for someone to comment this hah scared me too!
It was very very good so never stop.
this was helpful as a keystone to abstract algebra, thanks for the encouragement.
I am watching this in 2024 and it's very helpful.Thank you very much
The explanation was so simple and easy to understand.
Thank You !!!
At time mark 6:13, the elements of Z/3Z are cosets not numbers. The Cayley table is that of Z sub 3.
This channel is just wayy too good! :)
Love the vids! I’m binge watching the playlist before the algebra class next semester :D
Curiosity has me learning about octionions and above, this video is helpful in that endeavor
Very nice explanation
Legend in mathematics😍😍
i love the sound fx everytime there's a contradiction
I watched some of these for fun before. Now, I'm coming back to supplement the set theory in my discrete mathematics textbook.
5:59 symmetric (along the diag)how do you mean?🤔😐
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.
Thank you. This was an eye opener thought provoking video which cleared many of my doubts which I was searching for.
U
So beautiful explanation
I highly recommend reading Wikipedia and Proof-Wiki about Cayley tables for how they are used for non associative quasi-groups and other fun stuff.
thanks mam .your lecture is very helpful for me
You teaching style is awesome
Pls include a video on how to find the generators of a cyclic group of multiplicative order
Excellent.i learned very clearly algebra.
It's very helpful for everyone interested in mathematics.
Big fan of you... you explained very well❤❤
"3 groups are identical"
Isomorphic groups. Down the rabbit hole I go. If I weren't for searching for other materials, I would be stuck. Nonetheless, this is an excellent tutorial.
"2 distinct groups"
One for int mod 4, and one for that K4 group.
This was a very abstract excel tutorial
Thanks for all vedios you made, they are so exciting and easy to understand ❤❤
I am watching u r videos in 2020
COVID19
From kashmir
6:15 Kaylee tails and a comparison have a group with order three
Thank you for this video. It is really helpful
What a beautiful way to teach abstract algebra! Thanks a lot.
you are fun to watch, really you are doing a great job, abstract algebra was never fun. Thank you
2:00 Every row and column contains identity because,