Group Actions Part 4

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  • Опубліковано 18 гру 2024

КОМЕНТАРІ • 10

  • @Dummbeat
    @Dummbeat 8 років тому +6

    Extremely understandble explanation of the group actions here...Thank you very much for putting such effort into this. I finally understand the cocenpt of group actions and also the connection to the Cayleys Theorem.

  • @gutzimmumdo4910
    @gutzimmumdo4910 2 роки тому +2

    set permutations its just better specially with the venn function diagram representations with arrows.

  • @bujarshita148
    @bujarshita148 3 роки тому

    Good Work. Thank you. This nice explanation of group action should have at least one non-trivial example.

  • @darrenpeck156
    @darrenpeck156 2 роки тому

    What about structures for different kernals/quotient groups? What about typical homorphisms into set A?

  • @hyperduality2838
    @hyperduality2838 3 роки тому +1

    Group actions are dual to permutation representations!
    Injection is dual to surjection creates bijection or isomorphism.

  • @gavingordon7782
    @gavingordon7782 6 років тому +2

    Why is there a preference for using the group action method rather than the more intuitive permutation representations?

    • @MuffinsAPlenty
      @MuffinsAPlenty 4 роки тому +3

      Actions are the essence of scalar multiplication. For example, a vector space can be described as a group which has a field acting on it. The field forms the "scalars" and the group forms the "vectors". Sometimes the intuition coming from "scaling" is more useful than the intuition coming from permuting.

    • @hyperduality2838
      @hyperduality2838 3 роки тому

      Group actions are dual to permutation representations!
      Injection is dual to surjection creates bijection or isomorphism.
      Points are dual to lines -- the principle of duality in geometry.

  • @helloitsme7553
    @helloitsme7553 4 роки тому

    Can one imagine S_A as the set of all bijections from A to itself, even if A isnt finite?

    • @MuffinsAPlenty
      @MuffinsAPlenty 4 роки тому

      Yes. It makes perfect sense, even for infinite sets A.