Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a, a^2, a^3...a^{11}$?

share|cite|improve this question
See the OP's earlier question for context… – 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
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
up vote 3 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.

share|cite|improve this answer

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).

share|cite|improve this answer

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 $k=\frac{n}{\text{gcd}(n,i)}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.