group theory proof

Question

Let $G=\langle t \rangle$ be a cyclic group with $\text{ord}(t)=n$. I want to show that for all $d|n$ it holds that $$\left\lbrace s \in G ;\text{ord}(s)=d \right\rbrace=\left\lbrace t^{\frac{n}{d}k} ; 1 \leq k \leq d, 1=\text{gcd}(k,d) \right\rbrace $$

Answer 1

If you assume $m<d$ and $(t^{\frac{n}{d}k})^m=1$ it follows that $$t^{\frac{n}{d}km}=1 \Longleftrightarrow \frac{nkm}{d} = nz, z \in \mathbb{Z}\Longleftrightarrow \frac{km}{d} = z$$ which is a contradiction because $m<d$ and $\text{gcd}(k,d)=1$.

Answer 2

Let $G$ be cylic group with order $n$ with $G=\{e,t,t^2,...,t^{n-1}\}$.

fact$1$: $O(t^i)=\dfrac{n}{gcd(n,i)}$ where $O(t^i)$ represents the order of $t^i$.

By fact$1$ all generators of $G$ in the form $t^i$ where $(i,n)=1$.

fact$2$: For all $d|n$, Cyclic group has ecaxtly one subgroup of order $d$.

Now let $H$ be the subgroup of $G$ with order $d$. We know that it is also cyclic and since $a=t^{\frac n d}$ has order $d$ then $H=<a>$ since $H$ is uniqe .

By fact$1$, All generators of $H$ are of the form $a^i$ where $(i,d)=1$ we are done.