He does good work and gets good views for his content, the thing is that as farther in the world of pure math you get the less people can understand and the less views in general you will have as a CZcams instructor.
@@juniorochoazavalza228Well, not only pure math students watch these videos. I am physics student and I learn group theory and abstract algebra for its aplications in physics like in quantumn mechanics or QFT.
Hello sir, I tried to follow your steps with the group = Z4 (Integers mod 4), under multiplication. But I cannot verify that λg is injective (one-to-one). For eg. λ2(0)=2*0=λ2(2)=2*2=4=0. But 0 is not equal to 2 and thus the mapping is not injective. Can you pls help me with this confusion??
The Integers mod 4 is not a group under multiplication; there is no multiplicative inverse for 0 as no x*0 = 1 (the identity) in the set and every element must have an inverse for it to be a group. If you remove the element {0} it is still not a group as 2*2=0 which is not in the set so you wouldn't have Closure. The integers mod 5 (without 0) however would work.
@@mathmagic199 to make the group of the integers under multiplication you have to discludr all elements with no multiplicative inverses, all the elements that are not coprime to 4, so just 1 and 3
I assume you mean in my multivariable calculus videos? Once I finish all of the material associated with the current course I am teaching I plan to push into further directions -- general statements of vector calculus theorems in arbitrary dimensions and such.
@@MichaelPennMath I was thinking rather of establishing differential equations describing movements of particular systems, such as the double pendulum with 2 degrees of freedom, but perhaps this is no longer a matter of math but of physics
Can you show the permutations of this permutation group you name with Lambda? What are the actual permutations? And what groups are isomorphic? You keep naming groups without giving their elements, and using a bunch of variables and indexes as pseudo group elements, without giving the actual groups. What is the "group law" you mention and then don't define? Really hard to follow you if you don't define your objects
It depends on the chosen group. Whichever group you choose, its corresponding permutations are generated by the same operation which is followed by the group. For instance, in the example given in the video, the operation used to generate the permutations is of the group Z3, which is addition modulo 3. However, note that the permutations as a whole form a group under function composition. Summary of the story is that using the elements of a given group and its binary operation, you can produce a permutation group which is isomorphic to the original group.
Its so sad you dont get more views. You really teach outstandingly.
He does good work and gets good views for his content, the thing is that as farther in the world of pure math you get the less people can understand and the less views in general you will have as a CZcams instructor.
@@juniorochoazavalza228Well, not only pure math students watch these videos. I am physics student and I learn group theory and abstract algebra for its aplications in physics like in quantumn mechanics or QFT.
I want that blackboard erasing superpower...
Outstanding is indeed the best word to describe your teaching!
This make so much sense and meaning and do understand what to write
Best math CZcams channel
Excellent explanation,sir
Good explanation, i m Working on that theorem
Great explanation, thank you.
Thank you so much for this ❤️
really AWESOME!!!! THANKS a lot
Finally clear. You are epic!
Outstanding sir
thanks a lot sir!
Can you please do an example on how to find a subgroup of S6 that is isomorphic to D3?
nice video!
You're amazing
nice video
Hello sir, I tried to follow your steps with the group = Z4 (Integers mod 4), under multiplication. But I cannot verify that λg is injective (one-to-one). For eg. λ2(0)=2*0=λ2(2)=2*2=4=0. But 0 is not equal to 2 and thus the mapping is not injective. Can you pls help me with this confusion??
The Integers mod 4 is not a group under multiplication; there is no multiplicative inverse for 0 as no x*0 = 1 (the identity) in the set and every element must have an inverse for it to be a group. If you remove the element {0} it is still not a group as 2*2=0 which is not in the set so you wouldn't have Closure. The integers mod 5 (without 0) however would work.
@@thomaspickin9376 but in z4 not only 0 , the element 2 also has no inverse
@@mathmagic199 to make the group of the integers under multiplication you have to discludr all elements with no multiplicative inverses, all the elements that are not coprime to 4, so just 1 and 3
general question: do you intend to approach the generalized coordinates and to treat some cases of analytical mechanics ?
I assume you mean in my multivariable calculus videos? Once I finish all of the material associated with the current course I am teaching I plan to push into further directions -- general statements of vector calculus theorems in arbitrary dimensions and such.
@@MichaelPennMath
I was thinking rather of establishing differential equations describing movements of particular systems, such as the double pendulum with 2 degrees of freedom, but perhaps this is no longer a matter of math but of physics
If G=Z, the mapping x->10x won't surjective
Can you show the permutations of this permutation group you name with Lambda? What are the actual permutations? And what groups are isomorphic? You keep naming groups without giving their elements, and using a bunch of variables and indexes as pseudo group elements, without giving the actual groups. What is the "group law" you mention and then don't define? Really hard to follow you if you don't define your objects
It depends on the chosen group. Whichever group you choose, its corresponding permutations are generated by the same operation which is followed by the group. For instance, in the example given in the video, the operation used to generate the permutations is of the group Z3, which is addition modulo 3. However, note that the permutations as a whole form a group under function composition. Summary of the story is that using the elements of a given group and its binary operation, you can produce a permutation group which is isomorphic to the original group.
why showing that 𝜑(gh) equals λgh when proving isomorphic
To prove it is a group homomorphism