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Groups and subgroupd

  1. Nov 29, 2005 #1
    Let (G,*) be a group and (S,*) a subgroup of G. Prove that if for an element a in G, there exists m,n in Z, which are relatively prime, such that a^m and a^n is in S, then a is in S.

    At the moment, I think the problem is trivial but something just tells me it is not.
     
  2. jcsd
  3. Nov 30, 2005 #2
    What do you know about relatively prime integers? There's a well-known characterization of them which could be used here.
     
  4. Nov 30, 2005 #3
    So m and n have no common factors?
    Thus if a^m is in S and a^n is in S, a^m share no common divisors with a^n and vice versa so a must be in S?
     
  5. Nov 30, 2005 #4

    matt grime

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    It doesn't make sense to talk about divisors in a group like that. What does it mean to talk about common divisors of two rotations of a polygon?


    It isn't that m and n have no common factors: they do, all numbers have common factors, and one called the highest common factor springs to mind.
     
  6. Nov 30, 2005 #5
    Oh... that's what I meant. So if (m,n)=1, then only a could factors both a^n and a^m?
     
  7. Nov 30, 2005 #6

    matt grime

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    That still doesn't make any sense. Don't think in G or S for now.


    If I were to say: let m and n be coprime, then what must your first reaction be? (Don't mention G or S at all, please)
     
  8. Nov 30, 2005 #7
    Then the greatest common divisor for m and n is 1.
     
  9. Nov 30, 2005 #8

    matt grime

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    Well, obviously, since that is the definition of coprime, but what can you deduce from that (still ignoring G and S)? You have done Euclid's Algorithm?
     
  10. Nov 30, 2005 #9
    Yes, I've done Euclid's algorithm. So it states that dividing m and n over and over until there's a remainder that cannot be factored again will be the greatest common divisor. But since m and n are relatively prime, then I could find integers, p and q, such that mp + nq = 1?
     
  11. Nov 30, 2005 #10

    matt grime

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    Yes, and now what can you do with those when we allow ourselves to think about G and S and in particular a?
     
  12. Nov 30, 2005 #11
    Oh... I think I get it. Am I supposed to think in terms of a^mp + a^nq = a^1?
     
  13. Nov 30, 2005 #12

    matt grime

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    you can't add elements of the group like that. remember that multiplication of powers of an object adds the indices, ie (x^r)(x^s)=x^(r+s)
     
  14. Nov 30, 2005 #13
    So... if a^m is in S and a^n is in S, then neccessarily (a^m)(a^n) is in S which is equal to a^(m+n)?
     
  15. Nov 30, 2005 #14

    matt grime

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    yes, that is certainly true, but if x is in S, then x^p is also in S, isn't it, since S is a subgroup. I haven't picked p at random, and y^q would also be suggestively useful.
     
  16. Nov 30, 2005 #15
    Wow, thanks for all your help Matt.
     
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