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Let $G$ be a finite group of order $n$. If there are $e_d$ number of elements of order $d$ then the number of cyclic subgroups of order $d$ where $d$ is a positive divisor of $n$, is $\frac{e_d}{\phi(d)}$ where $\phi$ is Euler's phi function.

To show this we consider that there are $x$ number of cyclic subgroups of order $d$ and no two of them share common generators of order $d$. Since each subgroup will have $\phi(d)$ number of generators viz elements of order $d$ so $x\phi(d)=e_d$ i.e. $x=\frac{e_d}{\phi(d)}$.

I studied Gallian in which I found this argument. I do not recall the exact chapter but I think it was External direct product.

My question is the line where it was stated that no two cyclic subgroups of order $d$ will share any common generators. Why is this true ?How to prove this ?

Say $H, K$ be two cyclic subgroups of order $d$. How to show that $h$ and $K$ will share no common elements of order $d$ ? Did I frame the correct question or is it false in general ? If so what could be correct statement then ?

Any help please?

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    $\begingroup$ If $H$ and $K$ are (cyclic) of order $d$, and $x\in H\cap K$ has order $d$, then $H=\langle x\rangle = K$, because $\langle x\rangle$ is a subgroup of order $d$ of both $H$ and $K$. $\endgroup$ – Arturo Magidin Sep 5 at 16:39
  • $\begingroup$ @ArturoMagidin Got it. It was right before my eyes but could not see it...my bad :-D But anyway, thank you so much $\endgroup$ – Anjan3 Sep 5 at 18:53
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Since a group is cyclic if it is generated by one element... what can you say about two cyclic groups that are both generated by an element $x$?

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