Inner & Outer Semidirect Products Derivation - Group Theory

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  • Опубліковано 22 сер 2024

КОМЕНТАРІ • 20

  • @ProfOmarMath
    @ProfOmarMath 3 роки тому +23

    You've mathematically matured after a semester of college. It's really wonderful to see.

    • @MuPrimeMath
      @MuPrimeMath  3 роки тому +3

      Thank you for the kind words!

  • @user-xm9xo7jg4u
    @user-xm9xo7jg4u 3 роки тому +13

    This is the best lecture on semidirect products of groups I have ever seen. Thanks so much for your sharing.

  • @guohaoyang9163
    @guohaoyang9163 Рік тому +1

    This is the clearest video to introduce semi direct products! Thank you so much!🎉

  • @maximpushkar91
    @maximpushkar91 3 роки тому +6

    Very cool video! When this topic was on my lessons, I didn't really understand it. But now I can fully understand semidirect product, thank you!

  • @ahyunseo
    @ahyunseo 3 роки тому +5

    What a perfect lecture

  • @Astronautakaty
    @Astronautakaty 2 роки тому +2

    this is truly perfect, thank you so much for this clear lecture

  • @dipenganguly2635
    @dipenganguly2635 2 роки тому +2

    Nice video.. crystal clear

  • @ayabarre9315
    @ayabarre9315 Рік тому

    Thanks...we've never taken this idea before this way❤

  • @adresscenter
    @adresscenter 3 роки тому +2

    Very good teacher

  • @sbjnyc
    @sbjnyc 11 місяців тому +2

    I always wondered why the map is K-->Aut(H) rather than Inn(H) ... Can you make a semi direct product using outer automorphism?

  • @leeholzer4989
    @leeholzer4989 2 роки тому +1

    that was SO helpful

  • @artsygirlpoulami
    @artsygirlpoulami 3 роки тому +3

    Thanks for such a nice and methodical explanation. Can you suggest to me a book regarding this abstract group theory part? I am facing difficulty in my course work.

    • @MuPrimeMath
      @MuPrimeMath  3 роки тому +3

      The book that I used was Dummit and Foote. I didn't reference it very often so I don't actually know how helpful it is! However, it does have many group theory exercises for practice.

    • @artsygirlpoulami
      @artsygirlpoulami 3 роки тому

      @@MuPrimeMath Thanks a ton. I shall have a look.

  • @sambhusharma1436
    @sambhusharma1436 Рік тому

    Very nice 👍

  • @awaiskhan8327
    @awaiskhan8327 2 роки тому

    How do we know that h1 = h2 and k1 =k2 are we assuming that each element in the group G, H is it's own inverse ?

  • @tomkerruish2982
    @tomkerruish2982 3 роки тому +2

    Gotta say, you remind me of a young Allen Knutson, except that you're probably taller and I don't know if you juggle.

  • @alexandrafant5735
    @alexandrafant5735 10 місяців тому

    When you’re showing its a homomorphism, how is it not assuming the goal? I thought phi((h1,k1)(h2,k2))=phi((h1,k1))phi((h2,k2)) was the goal to show its a bijection, but in the end you use it as a fact to justify the products solution? I think I’m missing something

    • @alexandrafant5735
      @alexandrafant5735 10 місяців тому

      Ohhh nvm, we’re finding a law that makes it true! I get it now