Visual Group Theory, Lecture 3.6: Normalizers

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  • Опубліковано 7 сер 2024
  • Visual Group Theory, Lecture 3.6: Normalizers
    A subgroup H of G is normal if xH=Hx for all x in G. If H is not normal, then the normalizer is the set of elements for which xH=Hx. Obviously, the normalizer has to be at least H and at most G, and so in some sense, this is measuring "how close H is to being normal". We interpret this in terms of Cayley diagrams, and then prove some basic properties of normalizers: they are always subgroups, and they are unions of cosets -- precisely, those left cosets that are also right cosets.
    Course webpage (with lecture notes, HW, etc.): www.math.clemson.edu/~macaule/...

КОМЕНТАРІ • 16

  • @shaisimonson3330
    @shaisimonson3330 5 місяців тому +2

    Motivating the normalizer by quantifying the notion of normality through voting is a great pedagogical start.

  • @katyohsiek915
    @katyohsiek915 3 роки тому +3

    this lecture series is SAVING me!!! thank you so much for uploading these!

  • @pu5epx
    @pu5epx Рік тому

    Greetings from Brazil. This series is my favorite soap opera! :)

  • @ir6plans60
    @ir6plans60 7 місяців тому

    I love the opening question.

  • @manassrivastava1048
    @manassrivastava1048 2 роки тому +2

    You are awesome Professor! Even though my semester is over I am binge watching your videos instead of Netflix! Thanks a lot and looking forward to more amazing math! Is it possible for you to do a lecture series on representation theory, I am gonna take that course next sem and I am a final year undergrad student from India! Thanks a lot!

  • @wotanxiaozuo
    @wotanxiaozuo 7 років тому +1

    Very good explanation!

  • @nainamat6861
    @nainamat6861 2 роки тому

    Voting idea for explaining is perfect, thank you professor 🙏😊😃

  • @atzuras
    @atzuras 2 роки тому +1

    In every math course there's a point in which my neuron snap.
    no wait...
    yeah it's gone.

  • @maurocruz1824
    @maurocruz1824 2 роки тому

    13:10

  • @pawebielinski4903
    @pawebielinski4903 11 місяців тому

    I have a problem with the proof of observation 1. Why is the last equality (i. e. Hg=Hb) true? We only assumed that gH=bH, in other words b \in gH. For this equality we would need b \in Hg.

    • @pawebielinski4903
      @pawebielinski4903 11 місяців тому

      Ah I see, that's because gH=Hg so b\in gH is equivalent to b \in Hg. Ok

  • @smackronme
    @smackronme 7 років тому +2

    I guess the color of x at 8:59 should be blue..

    • @Mrpallekuling
      @Mrpallekuling 2 роки тому

      Yes, same error as in the previous lecture (3.5).

  • @ChuanChihChou
    @ChuanChihChou 2 місяці тому

    Gerrymandering of group theory

  • @ijindela5722
    @ijindela5722 7 років тому +1

    1:30 Should be "at minimum ONE element (e) votes "yes""

    • @deepaksachan5240
      @deepaksachan5240 5 років тому +6

      No, this is correct. Notice every g in H will always satisfy gH=Hg because H is a subgroup.