How can I find the number of elements of order $d$ (where $d$ divide $n$) in a cyclic group $G$ generated by $g$ of order $n$? I try to find all the elements $g^k\in G$ such that $o(g^k)=\frac{n}{gcd\{k,n\}}=d$ but in this way I have to test it for every $g\in G$. So my question is: is there a faster way to count it?
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Here are the key facts:
An element of order $d$ generates a cyclic subgroup of order $d$.
Every subgroup of a finite cyclic group of order $n$ is cyclic of order a divisor of $n$.
A cyclic group of order $n$ has exactly one subgroup of order $d$ for each divisor $d$ of $n$.
A cyclic group of order $d$ has $\phi(d)$ generators.
lhf
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Can you be more specific? – MayoDancer May 24 '18 at 14:10
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@MayoDancer, in a cyclic group of order $m$, there are exactly $\phi(m)$ generators. – lhf May 24 '18 at 14:36
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See also https://math.stackexchange.com/a/410464/589 – lhf May 24 '18 at 14:39