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I know the order of the group is the number of elements in the set. For example the group of $U_{10}$ (units of congruence class of 20) has order 4.

Major Edit, kinda changed the question. Lets say my element $a$ has a finite order $n$. Then what is the order of $a, a^2, a^3...a^{11}$?

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See the OP's earlier question for context math.stackexchange.com/questions/27129/… –  Jack Schmidt Mar 28 '11 at 22:27
    
Edited my question as Rasmus was right. I just read Wikipedia, but I am still very confused. –  Tyler Hilton Mar 28 '11 at 22:50
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based on this question and some of your previous related questions, I am concerned that you may be having difficulties beyond those which a math Q&A site can help you with. If you are taking a course, I strongly recommend that you talk to your instructor. If not, then I recommend that you find an actual "analog" person who can give you one-on-one assistance. –  Pete L. Clark Mar 29 '11 at 0:15

3 Answers 3

up vote 2 down vote accepted

Suppose that $\rm\:a\:$ has order $\rm\:n\:.\:$ To compute the order of $\rm\:a^i\:$ one may proceed efficiently as follows

$$\rm a^{i\:k} = 1\ \iff\ n\ |\ i\:k\ \iff\ n\ |\ i\:k,\:n\:k\ \iff\ n\ |\ (i\:k,n\:k) = (i,n)\:k\ \iff\ n/(i,n)\ |\ k$$

Therefore $\rm\:a^i\:$ has order $\rm\:n/(i,n)\:.\:$

Note especially how this method efficiently simultaneously proves both directions of the proof by exploiting the universal bidirectional $(\iff)$ definition of the $\rm gcd,$ namely $\rm\ a\ |\ b,c\ \iff\ a\ |\ (b,c)\:.\:$ Contrast this with standard proofs (e.g. other answer) which prove each direction separately.

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The order of an element in a group is the order of the subgroup it generates. Equivalently, it is the least integer n such that $a^n$ is the identity. If the order of the group is n, then the order of any element in the group actually divides n (this is Lagranges theorem).

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Answer to the edited question: That depends on n. Generally the order of $a^i$ is $\frac{n}{\text{gcd}(n,i)}$. If you want to prove this, you have to check two things:

Firstly that $(a^i)^{\frac{n}{\text{gcd}(n,i)}}=a^{\frac{in}{\text{gcd}(n,i)}}=1$. This holds because we have a multiple of n in the exponent and $a^{kn}=(a^n)^k=1^k=1$.

Secondly that this is in fact the smallest exponent k with $(a^i)^k=a^{ik}=1$. Since n is the smallest number with $a^n=1$, we are looking for the smallest number k so that $ik$ is a multiple of n. This is $k=\frac{n}{\text{gcd}(n,i)}$.

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Your final sentence requires proof. In fact it is the nontrivial direction of the theorem to be proved. To simply state that is true is begging the question. –  Bill Dubuque Mar 28 '11 at 23:49
    
you are right. One may show the last sentence by looking at the prime factorisation of i and n. –  Michalis Mar 29 '11 at 9:10

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