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Groups Theory Proof

  1. Feb 11, 2005 #1
    For a finite group G, I need to prove that the order of an element in G is a divisor of the order of the group. I'm not sure what this exaclty means, but I think you have to use cyclic groups such that G={a^0,a^1,.....,a^n) where n+1 is the order of the group and a^0 is the identity element. So I think I need to use Lagrage's thm stating that the order of a subgroup is a divisor of the order of the group. So to find a subgroup of G seems to be a problem. Could I use a^m as a subgroup, where m < n+1?? What is exaclty meant by order of the element. Thanks.
     
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  3. Feb 11, 2005 #2

    Hurkyl

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    The order of an element a of a group G is the minimum positive integer n such that an = &epsilon;.
     
  4. Feb 11, 2005 #3
    The elements 1,a,^2,...A^n=1 that Hurkyl mentions generate a group, generally a subgroup of a larger group. Perhaps the group consists of 1 or perhaps it consists of all elements, in either case it divides the order of the group. So, the final case is when the group generated by a is neither of order 1 or the order of the entire group.....
     
  5. Feb 12, 2005 #4

    mathwonk

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    this result seems basically to be due to gauss, who studied it in the case of modular integers.

    the idea is basically to let x be any element not among the powers {a, a^2,...,a^n = e}, and consider the products xa, xa^2,...,xa^n = x. and show none of these are among the elements {a, a^2,...,a^n = e} either.

    continuing this process, one fills up the group by disjoint translates of the subgroup {a, a^2,...,a^n = e}. qed.
     
    Last edited: Feb 12, 2005
  6. Feb 12, 2005 #5
    some books (hungerford's) say the order of an element is the size of the cyclic group generated by that element. other books say it's the smallest n such that for an element a, a^n = e

    i think it would follow from the above definition. the group generated by an element a (whose size = o(a) ) is a subgroup of the big group so the theorem follows from lagrange's theorem.
     
  7. Feb 12, 2005 #6

    mathwonk

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    i just sketched the proof of "la grange's theorem" and pointed out that gauss already proved it for modular integers hence generated the key idea, namely exhausting a group by disjoint translates. there is no other idea in la grange's result hence to me, he deserves no credit.

    of course it can be deduced by quoting "his" theorem.
     
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