I study abstract algebra for fun, and I gotta say, this content is extremely engaging despite it being the first time I hear about some of these concepts. This is great.
I am a student of pure mathematics, I heard about your channel and I started watching your videos around 1 hour ago or so.. and I am shocked at your way of delivering..I mean you are out of this world
I cannot thank you guys enough. What a lucid way of explaining things!!! Why can't all our teachers be like this? I came looking for just one concept in Abstract Algebra and I am just hooked to the series. After watching these videos sooo many topics are much clearer. Otherwise I would just rote learn stuff for my exams. Subscribed right away. I ll watch anything you teach
Don't be mad: I used this video series to go to sleep. It worked really well! Now I want to go back and hear all the parts I missed because the parts I listened to were terrific.
Thank you for creating this top quality series of videos, this is my third day with you guys, I have just finished the course learning Rings, and as I felt myself almost as a "Lord of the Rings" I needed a guide into the Kingdome of Groups. Thank you again talking about the magic of all those transformations using "student's" math language, instead of PhD's math language. The Groups and Rings are just groceries from the outdoor market and you definitely know how to cook and how to present all the "grocery" material in the way that it is (1) yummy, (2) easy to digest, (3) all flavors are blooming, (4) the more I eat the material the hangrier I get ---> it awakes the hanger for knowledge, not an every professor is capable to awake the hunger for the subject he/she teaches. You do!!!
To anyone who thought this was going to be a presentation about biking: cyc! But seriously though, these videos are phenomenal, and I’ve already started telling other people about them since I discovered them yesterday.
You cannot imagine how excellent your lessons are. As soon as I listen to the video and watch it, I find myself gaining great knowledge of the simplest way in which mathematics can be explained. Thank you so much so garateful for you دايما بشوف شروحات عاليوتيوب بس شروحاتك من الأكثر ابداع وتميز
0:00 What is permutation? 0:41 Symmetric Group & its operation 2:05 Do we have a short form of permutation? -> Cycle notion 9:19 Cycles can be commutative or not? Thanks for your explanation, so clear ❤
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@@SameerKhan-nd5qb In this case it's the smallest power of an element that yields the unit element of the group. So if you multiply the permutation (1 2 3) three times by itself you get the identity permutation e that does not permutate any elements, i.e. (1 2 3) (1 2 3) (1 2 3) = (1 2 3)^3 = (1) (2) (3) = e
Conjectures |( a b )| = 2 = 2 because there is 2 elements in the cycle, a maps to b and b maps to a |( a b c)| = 3 |( a b c d)| = 4 … |( 1 2 3 … n)| = n
YES, if you multiply (a,b) with itself twice you will get (e) the identity cycle, the same thing if you multiply (a,b,c) with itself three times and the same thing for (a,b,c,d) multiplied with itself four times.
@@humamalsebai I tried this with a=(1,3,2) and after calculating (a cube) I got a^3=(1)(2)(3) . Please correct me if I am wrong or there is something else to consider.
@@rishidusad2985 your calculation seems right It wasn't mentioned in the video, but if you think about it - what would be the identity element look like as a composition of cycles? Well, for Sn, it's (1)(2)(3)...(n), which means 1 maps to 1, 2 maps to 2 ... n maps to n (or that's the identity element) - you can clearly see it if you give yourself an example The way you power up cycles with no repetitions of elements is that you basicly 'jump' as many times as the power Example: (for Sn, where n = 3) The notation (1 3 2) [2] = 1 means where to map the element 2 (as you can see, element 2 is mapped to 1 in the cycle (1 3 2)) if we apply this 3 times in a row, we get: (1 3 2) [1] = 1 --> 1 'jumps forward' 3 times, meaning that 1 goes to 3, goes to 2, goes to 1, so 1 goes to 1 The same for (1 3 2) [2] = 2 and (1 3 2) [3] = 3 Result: 1 maps to 1, 2 maps to 2 and 3 maps to 3, so it's (1)(2)(3) Another interesting thing which requres proof but is always true is: For every cycle of length n, if you multiply it by itself k times, where k is such a number that: n=mk, where m is an integer (said othwerwide, k divides n), then the cycle is 'broken' into k independent cycles of length n/k (independent cycles meaning cycles with no common numbers) Example: in Sn, n = 6, if we power up the following cycle (6 4 3 5 1 2) by 2, we get the cycles (5 3 1) (4 5 2) if we power up the same cycle by 3, we get (6 3) (4 1) (3 2) And therefore, if we power up it by 6, the same rule applies, becuase 6 divides 6 and we break our cycle of length 6 into 6 cycles of length 1, which is in fact our identity element. EDIT for @Samuel Rho, when we map 1 element to itself, we get 1 cyce of length 1, so if a =(1,3,2), then a^3 is not (1,3,2), instead a^3 = (1)(2)(3) which is the identity element
Absolutely amazing!! Just found you're channel today!, and my gosh what a great find! You're videos are so inspiring , fascinating and well made!!please please keep it up!! You're so awesome!!!! :)
Let x be some n-cycle. Pick any element from x's cycle and call it x_1. By definition of an n-cycle, there are n elements such that x_1 maps to x_2, x_2 maps to x_3, etc., until x_n maps back to x_1, and furthermore, x_1,x_2,...,x_n are unique. Therefore: x^1 maps x_1 -> x_2 x^2 maps x_1 -> x_2 -> x_3 ... x^n maps x_1 -> x_2 -> ... -> x_n -> x_1 And since x^n maps each element in the cycle back to itself (since x_0 was arbitrary), it follows that x^n = 1. Therefore, the order of x is n. [Note: I assume that n>2 in the examples. But it is obvious that it n=2, then x^2 maps x_1->x_2->x_1, and so the order is 2.]
I'm in abstract algebra right now so I'll give the answer: lcm(cycle lengths) = order where lcm(a, b, c, ...) = lowest common multiple of a, b, c, ... An example from my class: Find a shuffle of 13 cards that takes exactly 20 shuffles to return to the original order. (an) Answer: (1 2 3 4 5) (6 7 8 9) (10 11 12 13) The order is 20 because lcm(4,5) = 20. There are many solutions, though. A good discrete math problem would be to figure out how many there are... Don't forget to consider solutions with 1-cycles or 2-cycles, as those are also possible
On the challenge: If cycles are just another notation for permutations, the order of a cycle stands for the order of an element of the permutation group, right? In that case, it would get n compositions to get the identity permutation. So the order of an n-cycle is equal to n? Please correct me, if there is something missing. I’m really eager to learn :)
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These video should be sponsored by department of education of the world.
right? The difference in quality between math youtubers and public universities is getting outrageous.
There is no departament of education of the World.
Unicef and Unesco should come forward. I am your learning partner. Thank you teacher!
If only that was the world we lived in
Literally
I study abstract algebra for fun, and I gotta say, this content is extremely engaging despite it being the first time I hear about some of these concepts. This is great.
We love to hear this!! 💜🦉 #LifelongLearningFTW
its equivalent non formal description might also be useful.
for fun? Damn we have different definitions of fun
You're actually amazing!!!! It's hard trying to understand in a lecture this topic and with the lecture notes too! Honestly, you're legendary!!!!
Exactly
One of the finest explanations of any topic on youtube. Congrats, and thank you
I am a student of pure mathematics, I heard about your channel and I started watching your videos around 1 hour ago or so.. and I am shocked at your way of delivering..I mean you are out of this world
Thank you so much for your Abstract Algebra playlist. I survived the semester because of you.
Congratulations on your hard work!! And thank you for telling us you found our videos helpful. It really motivates us to keep making videos!! 💜🦉
This was seriously a life saver! Clearest, simplest, and most applicable summary I've seen
Didn't cost $80k a year neither!
I cannot thank you guys enough. What a lucid way of explaining things!!! Why can't all our teachers be like this? I came looking for just one concept in Abstract Algebra and I am just hooked to the series. After watching these videos sooo many topics are much clearer. Otherwise I would just rote learn stuff for my exams. Subscribed right away. I ll watch anything you teach
Don't be mad: I used this video series to go to sleep. It worked really well! Now I want to go back and hear all the parts I missed because the parts I listened to were terrific.
We do the same thing when we find a comforting video! We're so glad you're watching. 💜🦉
Thank you for creating this top quality series of videos, this is my third day with you guys, I have just finished the course learning Rings, and as I felt myself almost as a "Lord of the Rings" I needed a guide into the Kingdome of Groups. Thank you again talking about the magic of all those transformations using "student's" math language, instead of PhD's math language. The Groups and Rings are just groceries from the outdoor market and you definitely know how to cook and how to present all the "grocery" material in the way that it is (1) yummy, (2) easy to digest, (3) all flavors are blooming, (4) the more I eat the material the hangrier I get ---> it awakes the hanger for knowledge, not an every professor is capable to awake the hunger for the subject he/she teaches. You do!!!
Facts
In any other formal lecture this would have been mysterious, very mysterious.
These are so great! Reading my textbook is very difficult because everything looks so jumbled but you make everything make sense!
this teacher and this material are ideal, I don't know what to say, it's just clearly understandable material, thank you
3:42 I felt sure she was going to say "tricycle" and "bicycle"
You're right. And she leaves off the unicycle.
That would have been grandiose
It makes me sad that that isn't the terminology.
To anyone who thought this was going to be a presentation about biking: cyc! But seriously though, these videos are phenomenal, and I’ve already started telling other people about them since I discovered them yesterday.
You cannot imagine how excellent your lessons are. As soon as I listen to the video and watch it, I find myself gaining great knowledge of the simplest way in which mathematics can be explained.
Thank you so much so garateful for you
دايما بشوف شروحات عاليوتيوب بس شروحاتك من الأكثر ابداع وتميز
0:00 What is permutation?
0:41 Symmetric Group & its operation
2:05 Do we have a short form of permutation? -> Cycle notion
9:19 Cycles can be commutative or not?
Thanks for your explanation, so clear ❤
Topic samajh aa gya fir bhi 3 din se roj 10-10 baar dekh rha hu video
This is the finest youtube channel.
Thank you very much
BLESS THIS CHANNEL!
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Bestest education tutorial i ever seen 🙏 love from India ...It's easiest to understand the whole concept just beacuse of you !! Keep it up
Keep at it Socratica! Love the technical level and animations you ladies provide! 🤩
Was like revisiting my older maths lessons. Great explanation ,very lucid
I study this in polish language, but still i understand more from your videos than on lectures. Great job
rel
the only video i needed for my exams
best videos on abstract algebra, really helpful for my gate preparation
Amazing explanation. Really helped me understand the fundamentals of this section in my cryptography class! Thank you
Thank you so much Ma'am, this is the best explanation I've seen on UA-cam
Wow! wan't expecting such a great video. Thousands of people will save millions of hours because of it. Thanks.
The order of an n-cycle should be n. And consequently the order of a permutation is the LCM of the lengths of all its cycles.
What does order mean?
@@SameerKhan-nd5qb the number of elements in a group
@@SameerKhan-nd5qb In this case it's the smallest power of an element that yields the unit element of the group. So if you multiply the permutation (1 2 3) three times by itself you get the identity permutation e that does not permutate any elements, i.e. (1 2 3) (1 2 3) (1 2 3) = (1 2 3)^3 = (1) (2) (3) = e
As always is a big pleasure to learn with you. Thank you a lot.
Conjectures
|( a b )| = 2 = 2 because there is 2 elements in the cycle, a maps to b and b maps to a
|( a b c)| = 3
|( a b c d)| = 4
…
|( 1 2 3 … n)| = n
came here to say this. thank you
Real thing is order of cycle is LCM of each cycle.
So LCM of 2, if there is 1 then it is 2.
And so on upto |n- cycle|
YES, if you multiply (a,b) with itself twice you will get (e) the identity cycle, the same thing if you multiply (a,b,c) with itself three times and the same thing for (a,b,c,d) multiplied with itself four times.
@@humamalsebai I tried this with a=(1,3,2) and after calculating (a cube) I got a^3=(1)(2)(3) . Please correct me if I am wrong or there is something else to consider.
@@rishidusad2985 your calculation seems right
It wasn't mentioned in the video, but if you think about it - what would be the identity element look like as a composition of cycles? Well, for Sn, it's (1)(2)(3)...(n), which means 1 maps to 1, 2 maps to 2 ... n maps to n (or that's the identity element) - you can clearly see it if you give yourself an example
The way you power up cycles with no repetitions of elements is that you basicly 'jump' as many times as the power
Example: (for Sn, where n = 3)
The notation (1 3 2) [2] = 1 means where to map the element 2 (as you can see, element 2 is mapped to 1 in the cycle (1 3 2))
if we apply this 3 times in a row, we get:
(1 3 2) [1] = 1 --> 1 'jumps forward' 3 times, meaning that 1 goes to 3, goes to 2, goes to 1, so 1 goes to 1
The same for (1 3 2) [2] = 2 and (1 3 2) [3] = 3
Result: 1 maps to 1, 2 maps to 2 and 3 maps to 3, so it's (1)(2)(3)
Another interesting thing which requres proof but is always true is:
For every cycle of length n, if you multiply it by itself k times, where k is such a number that: n=mk, where m is an integer (said othwerwide, k divides n), then the cycle is 'broken' into k independent cycles of length n/k (independent cycles meaning cycles with no common numbers)
Example: in Sn, n = 6, if we power up the following cycle (6 4 3 5 1 2) by 2, we get the cycles (5 3 1) (4 5 2)
if we power up the same cycle by 3, we get (6 3) (4 1) (3 2)
And therefore, if we power up it by 6, the same rule applies, becuase 6 divides 6 and we break our cycle of length 6 into 6 cycles of length 1, which is in fact our identity element.
EDIT for @Samuel Rho, when we map 1 element to itself, we get 1 cyce of length 1, so if a =(1,3,2), then a^3 is not (1,3,2), instead a^3 = (1)(2)(3) which is the identity element
Explanation is fantastic .no word to say that what's it helpful of every student.
Wow what a nice timing, my algebra 1 exam is thursday, thank you that was awesome!
just want to thank you from the bottom of my heart makes so much sense now = )
One of the best lecture series on abstract algebra!😍😍😍
such an underrated channel.
You're so kind. We're glad you're watching. 💜🦉
Thank you! I was super confused and this video cleared all my doubts!!!! I pressed the like button and sent this to my fellow colleagues
Ahhh thank you SO MUCH for sharing!! It makes a huge difference for us. We're so glad you found our video helpful! 💜🦉
Thank you very much. I struggled a lot to understand this. You are really a great teacher. All the best. I hope more great contents are coming.
Very nice video. Was not only educational but really relaxing. Very simple explanation, great job!! 👌
the best channel for mathematicians
best of luck
Great explanation. It helped me in one of my assignments.
Can't believe it took me a long time to understand this.
Thank you, u've made each topic so clear in a simplified method, u are great! I love the way u explain everything
You make concepts easy!
Absolutely amazing!! Just found you're channel today!, and my gosh what a great find! You're videos are so inspiring , fascinating and well made!!please please keep it up!! You're so awesome!!!! :)
Thanks!
Thank you so much for your kind support!! 💜🦉
Great socratica. I am watching this after 4 year but still I can understand this... Great way of teaching.. thank you ☺️
so amazing way of teaching . i am watching this from india .keep doing up,best of luck for your faculties for making such a amazing animation ......
I never expect math can be that clearly explained
I'm in math prep school and we just saw this in class a few weaks ago ! Neat :)
Thank you so much. The video made me understand my lecture notes I had written without having an idea of.
SAME
Let x be some n-cycle. Pick any element from x's cycle and call it x_1. By definition of an n-cycle, there are n elements such that x_1 maps to x_2, x_2 maps to x_3, etc., until x_n maps back to x_1, and furthermore, x_1,x_2,...,x_n are unique. Therefore:
x^1 maps x_1 -> x_2
x^2 maps x_1 -> x_2 -> x_3
...
x^n maps x_1 -> x_2 -> ... -> x_n -> x_1
And since x^n maps each element in the cycle back to itself (since x_0 was arbitrary), it follows that x^n = 1. Therefore, the order of x is n.
[Note: I assume that n>2 in the examples. But it is obvious that it n=2, then x^2 maps x_1->x_2->x_1, and so the order is 2.]
Bravo!
Thank you
the music in the beginning is like from Duna
awesome, it's like I'm learning to live in the sands world
*Thank you, this is really useful for me!*
Excellent explanation of the concept. Learnt after a long time.
I wish you could be my tutor. The way you analyse and sort things are outstanding..
When the dramatic music started I felt as if I am in a Christopher Nolan film.
This is EXACTLY what I needed!😭 Thank you!
thank you for this video. helped me a lot before an exam.
Liliana, muito obrigado pelo seu trabalho!
Feliz dia dos professores!!
Appreciated this video feel super ready for my exams now! i lied just got whole lot more confused about cycle composition.
Awesome!! Thank you, this is a wonderful approach to dealing with a tricky situation.
You make math super easy!!!
Mam I like very much your way of teaching, thank you very much for presenting this
This was actually so pretty amazing. I just can't praise enough.
I really would love this video 4 months ago when I was stuying this, nice vid btw
Thanks for making this so easy to understand
I'm in abstract algebra right now so I'll give the answer:
lcm(cycle lengths) = order
where lcm(a, b, c, ...) = lowest common multiple of a, b, c, ...
An example from my class:
Find a shuffle of 13 cards that takes exactly 20 shuffles to return to the original order.
(an) Answer:
(1 2 3 4 5) (6 7 8 9) (10 11 12 13)
The order is 20 because lcm(4,5) = 20.
There are many solutions, though. A good discrete math problem would be to figure out how many there are...
Don't forget to consider solutions with 1-cycles or 2-cycles, as those are also possible
Thanks for the explanation!
Could you please do a video on orbits and stabilisers, you’re great at explaining abstract algebra and I can’t find any videos on this that are good
Thanks from🇵🇰
And a lot of respect
Do more in this field
This was extremely helpful. Thank you so much!
Nice explanation. Cristal clear!
Wow! So clear explanation. Thank you.
Your way of teaching 👌👌awsmm
Amezingggg way to explain my concept is completely clear ❤after seeing this vedio
Cheer~~~a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.😊
thank you for such simple and easy explanation making math fun!!!! 😀
So great!
Great tutorial! Thanks a lot!
bah!!! vaba jay na !! darun! kudos
melhor explicação sobre a temática que já vi.....
Thank you so much😭💕such a clarifying explanation 😊
Thank you so so so much for this great video. It has helped me a lot. Best regards and wishes to you
On the challenge:
If cycles are just another notation for permutations, the order of a cycle stands for the order of an element of the permutation group, right? In that case, it would get n compositions to get the identity permutation. So the order of an n-cycle is equal to n?
Please correct me, if there is something missing. I’m really eager to learn :)
Yes, the order of an n-cycle is n. It generates a subgroup of S(n) isomorphic to Z(n) (aka Z/nZ).
@@shacharh5470 Can you explain how?
You're really great 🥰 you made it so easy to learn ❤️
We're so glad you're learning with us!! 💜🦉
Where was that 1 semester ago? :(
I love this!
Great explanation! Really helpful.
An extraordinary work!!!
A lot of effort in these videos! really cool
Excelente vídeo minha amiga. Sempre com vídeos sensacionais!
amazing. i like how you explain dan the animation is very excellent.
Excellent video . Thanks
The video is very well made, Thank you.
thank you for these great videos!
This was very useful!
Amazing video, many thanks
simple and clear, have a like