Visual Group Theory, Lecture 5.4: Fixed points and Cauchy's theorem

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  • เผยแพร่เมื่อ 1 ก.พ. 2025

ความคิดเห็น • 14

  • @fsaldan1
    @fsaldan1 4 ปีที่แล้ว +3

    I missed where the Lemma is used in the proof of Cauchy's Theorem.

  • @sergiopellitero4136
    @sergiopellitero4136 20 วันที่ผ่านมา

    20:40
    Why |Orb(s)| can't be grater than p?

  • @sajateacher
    @sajateacher 3 ปีที่แล้ว +2

    Isn't there another Cauchy's theorem from Complex Analysis?

    • @alidursun1706
      @alidursun1706 3 ปีที่แล้ว +3

      of course, Cauchy is everywhere!

    • @Mrpallekuling
      @Mrpallekuling ปีที่แล้ว +1

      Here are some:
      Cauchy's integral theorem
      Cauchy's limit theorem
      Cauchy's residue theorem
      Cauchy's mean-value theorem in real analysis
      Cauchy's theorem on the rigidity of convex polytopes
      Cauchy-Hardemard theorem in complex analysis
      Cauchy-Peano theorem for ordinary differential equations
      Cauchy-Goursat theorem for complex-valued functions
      Cauchy-Kovalevskaya theorem about partial differential equations

  • @AlexTheProMacGamer
    @AlexTheProMacGamer 4 ปีที่แล้ว

    8:18 Where does the `k` comes from? p divides |G|, therefore |G|=pm for some m. Why is there an exponent?

    • @mzg147
      @mzg147 2 ปีที่แล้ว

      p can divide m too. If so, then |G|=p^2 m' for some m'. So you divide out p as long as it possible, i.e. m is not divisible by p. You get p^k m, for some k which may be 1, but you know that m is not divisible by p.

  • @renushekhawat4061
    @renushekhawat4061 6 ปีที่แล้ว

    Sir , What about the lemma ( near about 2.30) if |S| = p?
    Then |S|(mod p) = 0 so what does it tell? There is no fixed point in S? Or that all the p points of S are fixed?
    Or what can we say when |S|=p^k?

    • @sergiopellitero4136
      @sergiopellitero4136 20 วันที่ผ่านมา

      |Fix(phi)|=0 means no fixed points exists.

  • @thephysicistcuber175
    @thephysicistcuber175 7 ปีที่แล้ว +4

    2:48 p=6?

    • @dominikchmura5103
      @dominikchmura5103 4 ปีที่แล้ว

      By assumption, p must be prime

    • @thephysicistcuber175
      @thephysicistcuber175 4 ปีที่แล้ว

      @@dominikchmura5103 I'm not sure, but that might have been a joke on the fact that as was drawn p was 6.

  • @addemfrench
    @addemfrench 7 ปีที่แล้ว

    Wait, you say there are only two groups of order 6, but what about Z6?

    • @sebastien3411
      @sebastien3411 7 ปีที่แล้ว +2

      The "up to isomorphism" part is important. One of the two groups he shows is isomorphic to Z6.
      Note that Z/2 X Z/3 is cyclic and can be generated by (1,1) for instance. So it is isomorphic to Z/6