# Groups Theory Proof

PhysKid24
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.

## Answers and Replies

Staff Emeritus
Gold Member
The order of an element a of a group G is the minimum positive integer n such that an = &epsilon;.

robert Ihnot
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.....

Homework Helper
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:
fourier jr
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.