Groups Theory Proof

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  • #1
PhysKid24
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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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  • #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;.
 
  • #3
robert Ihnot
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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.....
 
  • #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.
 
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  • #5
fourier jr
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PhysKid24 said:
What is exaclty meant by order of the element. Thanks.
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

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??
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.
 
  • #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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