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Group Theory

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    If ##|G| = p^2## where ##p## is a prime, then ##G## is abelian.


    2. The attempt at a solution

    The book my proof gives is confusing at the very last part.

    Suppose ##|Z(G)| = p##. Let ##a\in G##, ##a\notin Z(G)##. Thus ##|N(a)| > p##, yet by Lagrange ##|N(a)| \ | \ |G| = p^2##.

    Why is ##|N(a)| \ | \ |G| = p^2##? Since ##|G| = p^2## and ##|N(a)|> p ## how does that make ##|N(a)| \ | \ |G| = p^2##?
     
  2. jcsd
  3. Nov 13, 2013 #2

    Dick

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    You are reading it wrong. |N(a)| | |G| just means that the number |N(a)| divides the number |G|. It doesn't have a numerical value. They are just saying |N(a)|>p and yet |N(a)| must divide |G|=p^2. So?
     
  4. Nov 13, 2013 #3
    So, since ##|N(a)|>p## and ##|N(a)|## must divide ##|G|## then ##|N(a)| = p^2##?
     
  5. Nov 13, 2013 #4

    Dick

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    Yes!
     
  6. Nov 13, 2013 #5
    Thanks! Last question, since ##|N(a)| = p^2## why would this imply ##a\in Z(G)##?
     
  7. Nov 13, 2013 #6

    Dick

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    N(a) is the normalizer subgroup of a. If N(a)=p^2 then N(a)=G, right? You tell me why that should mean ##a\in Z(G)##.
     
  8. Nov 13, 2013 #7
    Since N(a) = G then N(a) generates the group G thus ##a\in Z(G)##?
     
  9. Nov 13, 2013 #8

    Dick

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    I don't think that really means anything. What does N(a) mean?
     
  10. Nov 13, 2013 #9
    It is the normalizer, so ##N(a) = \{ x \in G \ | \ xa = ax\}.##
     
  11. Nov 13, 2013 #10

    Dick

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    So in words, N(a) is the set of all elements of G that commute with a. Now what does ##a\in Z(G)## mean in words describing what a commutes with?
     
  12. Nov 13, 2013 #11
    The center of the group G is the set of elements that commute with every element of G. Where N(a) is the set of all elements of G that commute with a. So is it because then that will make G abelian hence G = Z(G) hence a in Z(G)?
     
  13. Nov 13, 2013 #12

    Dick

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    It is going to be shown abelian eventually, but you aren't there yet, so you can't use 'abelian' to prove a is in Z(G). First just tell me why a is in Z(G). Think about the words and use that N(a)=G (why do we know that?).
     
    Last edited: Nov 13, 2013
  14. Nov 13, 2013 #13
    Since N(a) = G then all elements of G commute with a thus why ##a \in Z(G)?##
     
  15. Nov 13, 2013 #14

    Dick

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    I'd feel more confident you understood that if you wouldn't end all of your comments in a '?'. If you remind me how we know N(a)=G that might restore my confidence. No '?', ok?
     
  16. Nov 13, 2013 #15
    Lol I understand. We got ##N(a) = G## since we establish that since ##|N(a)|>p## and ##|N(a)|## must divide ##|G|## then ##|N(a)| = p^2## since ##p^2 = |G|## then ##|N(a)| = |G|##.
     
  17. Nov 13, 2013 #16

    Dick

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    Right! N(a) is a subgroup of G having p^2 elements. But G has only p^2 elements. So |N(a)|=|G| means N(a) must equal G, which should be your final conclusion. Thinking about these things in words and using less symbols often helps.
     
  18. Nov 13, 2013 #17
    Yes, it does! Thank you very much!
     
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