Abstract Algebra | Subgroups of Cyclic Groups

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  • Опубліковано 11 вер 2024
  • We prove that all subgroups of cyclic groups are themselves cyclic.
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КОМЕНТАРІ • 9

  • @dharmapatel211
    @dharmapatel211 4 роки тому +10

    At 2:50, the conditions on the remainder should be 0

    • @incoherentlogic228
      @incoherentlogic228 4 роки тому

      No, this is correct. If your r = m, that means that you divided by m and got a remainder of m, which means you could have divided once more to get a remainder of 0.

    • @user-hq5fn6yv2v
      @user-hq5fn6yv2v 4 роки тому +5

      @@incoherentlogic228 No, this is not correct. You are right that r = m cannot happen, but if 0

  • @eukleidesk6759
    @eukleidesk6759 3 роки тому +3

    Don’t we need to first show S is not empty to get a minimal element? It’s easy if g has finite order, since |g| is in S, but I don’t see any obvious choice if g has infinite order.

  • @omargaber3122
    @omargaber3122 4 роки тому +4

    I don't know how to start, but let me explain it to you
    I graduated from the College of Science, Department of Physics and Mathematics and I was the first in my class, and currently I am in Russia trying to study masters, but things here are bad, so I try to search for a free scholarships or any other work ,,, I searched through many websites but no avail.
    I am talking to you because you seem to know many solutions,
    Please answer me positively or negatively
    thanks for your time Dr.Michael

    • @totatota4804
      @totatota4804 4 роки тому +1

      he is ordinary person, no one will help you 😂😂😂

    • @MichaelPennMath
      @MichaelPennMath  4 роки тому +2

      Unfortunately I am not really sure I can offer you any advice that you don't already know. All major Universities in the US offer teaching assistantships and/or research fellowships for Ph.D. students. These vary greatly in their level of competitiveness. I don't know much about the European System, but I think most EU Universities also offer funding for Ph.D. students.

  • @skatethe4881
    @skatethe4881 Рік тому +1

    "That's a good place to end this vi-"