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- Thread starter PhysKid24
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Hurkyl

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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.

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.

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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 = ePhysKid24 said:What is exaclty meant by order of the element. Thanks.

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.PhysKid24 said: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??

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mathwonk

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

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