Cyclic group 60

Question

In a cyclic group of order 60 find the elements of order 12.

then find the number of element that satisfy $x^{12}=e$ So if $x^3=e$ then $x^{12}=e$ And I know $x=e$. what next do I do?

Finally find all the elements that satisfy $x^{20}=e$. Can you make a conjecture about the number of elements that will satisfy $x^n=e$ in a cyclic group? I am completely stuck on this.

Answer 1

In general, if $g$ has order $n$, $g^k$ has order $n/(n,k)$. Can you prove this? Can you use this to solve your problem?

Answer 2

If $G$ is a cyclic group of order $n$ and $d$ is a +ive divisor of n.Then it has $|U(n)|$ elements of order d. Thus if group of order 60 is as $G=<x>$, then $x^5,x^{25},x^{35},x^{55}$ are the element of order $12$.

Answer 3

$x^5$ works, because $(x^5)^{12}=e.$ To find the others, you need to look at all of the numbers that are relatively prime to 12 - they are 5, 7, 11. Therefore, $x^5,(x^5)^5,(x^5)^7,(x^5)^{11}$ work, giving you the other answer above. There are 4 because $\varphi(12)=4.$