# Show no two cyclic subgroups of $G$ will share common generators.

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?

• 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 $K$. – Arturo Magidin Sep 5 at 16:39
• @ArturoMagidin Got it. It was right before my eyes but could not see it...my bad :-D But anyway, thank you so much – Anjan3 Sep 5 at 18:53

## 1 Answer

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 $$x$$?