Computing the order of a group

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Homework Statement



Let a be an element of a group an let |a| = 15. Compute the orders of the following elements of G

a) a^3, a^6, a^9, a^12

Homework Equations





The Attempt at a Solution



for the first part of part a, would a^3 be <a^3>=<e,a^3,a^6,a^9,a^12,a^15,a^18,a^21,a^24,a^27,a^30, a^33,a^36,a^39,a^42>
 
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|a|=15 means that 15 is the lowest power of a that is equal to the identity.

So, for example,
a^{42}=a^{15}\times a^{15} \times a^{12}=e \times e \times a^{12}=a^{12}
so you've got too many things in your list.
 
What I want to know is why the OP stopped after a^42 in particular. I mean going beyond a^15 is clearly wrong, but why stop at a^42? Is that the 15th power of a^3? I think so, from quickly scanning the list.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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