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The group is given as a multiplicative group of order 15 (with 15 elements).
So let's keep it multiplicative.
Every element x has order (least power = to identity e) which divides 15.
So order of x is 1, 3, 5, 15. If order is 1, 5 or 15 then {x^3,x^5,x^{9}}
= {e}, {x^3, e, x^4} or {x^3, x^5, x^9} all distinct in last two cases.
So, since the set has two distinct elements, the order of x is 3.
Then x^3=x^9=e and x^13=(x^12)x =ex=x. So x^{13n} = x^n. Since
x has order 3, x^{13n} achieves 3 values as n goes from 1 to infinity.
So {x^{13n} | n is in N} has 3 elements.
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