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I found this formula to determine the number of cyclic subgroups of order $k$ in a group $G$:

$$\frac{\text{number of elements of order k in G}}{\text{total numbers coprime to k}}. $$ so the number of cyclic subgroups of order $5$ in $S_5 = 24/4 = 6$.

The problem is how this formula is derived? If we divide number of elements of order $k$ by $\varphi(k)$, why do we get total number cyclic subgroups of order $k$?

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Consider a cyclic subgroup of order $k$ generated by an element $a$. Then the group is generated precisely by its elements $a^j$ with $j$ coprime to $k$. Thus there are $\varphi(k)$ elements that generate it. Thus when you count the elements of order $k$ in the group, you're counting each cyclic subgroup of order $k$ $\varphi(k)$ times, so you need to divide that count by $\varphi(k)$ to get the number of cyclic subgroups of order $k$.

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  • $\begingroup$ Thanks, I got it. $\endgroup$ – Mathsaddict Sep 10 '18 at 6:49

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