+liam cottrell Thanks for your feedback Iiam. I wrote this note below your comment but in case you didn't see it. - If you go to my UoH Algebra playlist you'll find a whole set of videos on abstract algebra and a few on linear algebra. They follow the text written by Joseph Gallian. I'm teaching at the University of Hyderabad in India this semester and made the videos for my students.
the order of (13)(254). Because this is distinct, you can say let a=(13)(254) then the order of a is just the lowest common multiple of the cardinality of the first cycle (13) which is 2 and the cardinality of the second cycle (254) which is 3, ie LCM(2,3)=6. Thank you
Thanks for your feedback Casper. If you go to my UoH Algebra playlist you'll find a whole set of videos on abstract algebra. They follow the text written by Joseph Gallian. I'm teaching at the University of Hyderabad in India this semester and made the videos for my students.
sorry to be so offtopic but does anybody know of a trick to get back into an instagram account? I somehow lost my password. I love any help you can give me!
Hello! I have a question that appeared in one test. We have 2 permutations, a = (2 5) (4 1 5) (2 5) and b = (3 7) (6 3 7), and these are both permutations of S7. I know that a = (1 2 4) but how do I write b in the simplest form? I dont have a '1' to start, so I dont know how to start to write the simple form. Hope you can answer! Good video, nice explanation!
You don't need to start with 1. The permutation b fixes 1. b(1) = 1. So start with 3 or 6 or 7. You will get (67). I usually start with 1 if it's not fixed, but that's not important. If you had started your cycle with 4 you'd get a = (412). Same, right?
+Christina Carter Hi great lecture ! the link about "the cycle notation" is not working , can you please post it? it is crucial to understanding the lecture better . thanks in advance !
Yes! In the video I think I did it like this (1234) = (14)(13)(12). Your way will always work though, to write permutations as products of transpositions.
Thank you for the conformation. The reason I asked is because I wanted to confirm an error in another youtube user's video on alternating groups. He claimed that (12)(23)(34) = (1432). I hope you continue making videos on higher mathematics; they are greatly appreciated.
I'm not sure what you are referring to, but (235)(124) should be simplified to (13524). Always leave them as a product of disjoint cycles. Then the order is the lcm of the cycle lengths.
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Really appreciate the video, having someone walk through examples like you did was really useful. Thanks!
+liam cottrell Thanks for your feedback Iiam. I wrote this note below your comment but in case you didn't see it. - If you go to my UoH Algebra playlist you'll find a whole set of videos on abstract algebra and a few on linear algebra. They follow the text written by Joseph Gallian. I'm teaching at the University of Hyderabad in India this semester and made the videos for my students.
the order of (13)(254). Because this is distinct, you can say let a=(13)(254) then the order of a is just the lowest common multiple of the cardinality of the first cycle (13) which is 2 and the cardinality of the second cycle (254) which is 3, ie LCM(2,3)=6. Thank you
Thank you Christina, that video cleard up a number of points with which I was confused.
Why we start with 1 when multiplying permutations? Thanks for answers.
Thanks for your feedback Casper. If you go to my UoH Algebra playlist you'll find a whole set of videos on abstract algebra. They follow the text written by Joseph Gallian. I'm teaching at the University of Hyderabad in India this semester and made the videos for my students.
Where can I see your remaining lectures
sorry to be so offtopic but does anybody know of a trick to get back into an instagram account?
I somehow lost my password. I love any help you can give me!
Hi teacher I want your number in the WhatsApp plz send me
I have a lot of math work and linear algebra to do and I will pay you for it
excellent video, thank you so much!!
A very helpful video, thanks alot
thnx it cover all the things regarding permutation of group..
Thanks a lot, please make more videos!
Really good instructional video, it helped me alot. Especially since the book we use in my lectures feels really unreadable for me.
Well explained, really appreciate it. Please I would like to see all video related to abstract algebra
Thanks Amma. Keep uploading more videos like this
Soo good thank u
Thanks a lot for the video. it helps a lot.... just to know if you have some video on sifting techniques.
Sorry I don't Younoussa.
Wow...well explained. Thanks so much.
Just great. THANK YOU
Hello! I have a question that appeared in one test.
We have 2 permutations, a = (2 5) (4 1 5) (2 5) and b = (3 7) (6 3 7), and these are both permutations of S7.
I know that a = (1 2 4) but how do I write b in the simplest form? I dont have a '1' to start, so I dont know how to start to write the simple form.
Hope you can answer! Good video, nice explanation!
You don't need to start with 1. The permutation b fixes 1. b(1) = 1. So start with 3 or 6 or 7. You will get (67). I usually start with 1 if it's not fixed, but that's not important. If you had started your cycle with 4 you'd get a = (412). Same, right?
Excellent explanation , thanks!
In french we call a one to one function a bijective function.
thanks it helped me a lot
Thank u mam it's helpful for DU students also.😀
Can anyone explain to me what is a permutation symmetry?
sorry .the Link to the slide where you do cycle notation ?
Sorry, I didn't explain cycle notation on a video. Here is a link to a nice article on it. www.cems.uvm.edu/~jwsands/251f13/cycle.pdf
+Christina Carter
Hi great lecture ! the link about "the cycle notation" is not working , can you please post it? it is crucial to understanding the lecture better . thanks in advance !
+frank256256
Here are two links: th-cam.com/video/lSaSiBzVFSI/w-d-xo.html and th-cam.com/video/gGir2mwNrbo/w-d-xo.html
+Christina Carter thanks ! i wanted your awesome style but ill watch those !
thanks alot again .
You are great baby...
yes baby thank so much u the best girllll
Pls make more videos mam...Ur explained very good
Very clear explanations thank you
Excellent explanation.
It help me a lot
Is this correct...
(12)(23)(34) = (1234)?
I just want to make sure I'm doing it right.
Yes! In the video I think I did it like this (1234) = (14)(13)(12). Your way will always work though, to write permutations as products of transpositions.
Thank you for the conformation. The reason I asked is because I wanted to confirm an error in another youtube user's video on alternating groups. He claimed that (12)(23)(34) = (1432).
I hope you continue making videos on higher mathematics; they are greatly appreciated.
no. Those are inverses.
Thanks a lot 🙏🙏🙏
really nice tutorial
perfect.
thank you !
What I came here for: 12:24 :)
Da budeš Mastaaa
Bidnem
Thanks ma'am 🙏🙏
It's easy to take it as (235)(124)
I'm not sure what you are referring to, but (235)(124) should be simplified to (13524). Always leave them as a product of disjoint cycles. Then the order is the lcm of the cycle lengths.
very helpful thanks
How can I show that A8 contains an element of order 15?
Justin arocho (123)(45678) has order 15 - the least common multiple of 3 and 5.