Extremely helpful video!!! Very intuitive. For all who are watching this without refreshed knowledge about group actions and orbit-stabilizer theorem: Watch his videos about these theorems first and then continue with this one!
Thanks for the tip. He lost me on my first viewing as I hadn't seen group actions/OS theorem. They are very very well explained and I come back to this video and now I can say that I pretty much understand the proof of Sylow's first theorem. Cool.
We talked bout Orbit and Stabilizers of an element in a set A, we never did talk about a stabilizer of set in a set, did we? Anyway good to know it works about the same way
Also, once you have found a subgroup of order p^\alpha, how do you find a subgroup of order p^/beta, where beta < alpha? Sylow's first theorem says that there is a subgroup of all powers of p.
Can someone help me understand 1 point. I went back and watched the Orbit stabilizer theorem videos. Nowhere goes it guarantee us that a Stabilizer exists for a particular size. In this proof we choose a set of size p^alpha. However, how can we be sure that a stabilizer group exists for p^alpha?
Video on Group Actions Part 1 around 30mins in; 2nd axiom of group actions about the identity element of group says e.a = a. This guarantees the existence of a stabilizer with at least one element for any group action by definition.
A set of any size or order (assume n) has a Stablizer as a subgroup of the larger Symmetric Group Sn. In the videos of Orbit Stabilizer Theorem - he was distinctly discussing about groupactions onto Sets A, and not group.
I do not understand sth. Lets take G as Z_12 which is of order 2^2 * 3. And if we take C={0,3,6,9} as the subset of size 2^2=4. And apply the group action g.C as 4.C =3.{0,3,6,9}={0,0,0,0}={0} . The result is not a 4 element subset. .Hence it is not a group action. What is wrong with that example?
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Extremely helpful video!!! Very intuitive. For all who are watching this without refreshed knowledge about group actions and orbit-stabilizer theorem: Watch his videos about these theorems first and then continue with this one!
Thanks for the tip. He lost me on my first viewing as I hadn't seen group actions/OS theorem. They are very very well explained and I come back to this video and now I can say that I pretty much understand the proof of Sylow's first theorem. Cool.
It is excellent..... you have done it really nice...............
Beautiful! Thank you!
Nice videos. Thank you for making them. How many different highlighter colors do you have?
beautiful proof!
We talked bout Orbit and Stabilizers of an element in a set A, we never did talk about a stabilizer of set in a set, did we? Anyway good to know it works about the same way
👍😊
Explanations are very explicit .
Also, once you have found a subgroup of order p^\alpha, how do you find a subgroup of order p^/beta, where beta < alpha? Sylow's first theorem says that there is a subgroup of all powers of p.
Can someone help me understand 1 point. I went back and watched the Orbit stabilizer theorem videos. Nowhere goes it guarantee us that a Stabilizer exists for a particular size. In this proof we choose a set of size p^alpha. However, how can we be sure that a stabilizer group exists for p^alpha?
Video on Group Actions Part 1 around 30mins in; 2nd axiom of group actions about the identity element of group says e.a = a. This guarantees the existence of a stabilizer with at least one element for any group action by definition.
A set of any size or order (assume n) has a Stablizer as a subgroup of the larger Symmetric Group Sn. In the videos of Orbit Stabilizer Theorem - he was distinctly discussing about groupactions onto Sets A, and not group.
What book is being used?
I do not understand sth. Lets take G as Z_12 which is of order 2^2 * 3. And if we take C={0,3,6,9} as the subset of size 2^2=4. And apply the group action g.C as 4.C =3.{0,3,6,9}={0,0,0,0}={0} . The result is not a 4 element subset. .Hence it is not a group action. What is wrong with that example?
The binary operation of the group Z_12 is the sum, not the product. The action gives you 4+C={4,7,10,1}