The following can be stated in many different ways. > [!NOTE] Lemma > Let $n\in \mathbb{N}$ and $C_{n}=\langle a \rangle$ denote the [[Cyclic Group|cyclic group]] of order $n$. Let $d\mid n$. Then $C_{n}$ has a **unique** subgroup of order $d$ namely $\langle a^{n/d} \rangle$. **Proof**: Suppose $C_{n} = \langle a \rangle$. Then By [[Order of Power of Group Element]] $\text{ord}(a^{n/d})=d$. Then, $H_d=\left\langle a^{\frac{n}{d}}\right\rangle$ is a subgroup of $G$ of order $d$. (Uniqueness): Suppose, $H \leq G$ and $|H|=d$. We claim that $H=H_d$. It is enough to prove one-sided set inclusion $H \subseteq H_d$ because $H \leq H_d$ and $|H|=\left|H_d\right|$ implies $H=H_d$. Choose $b \in H$. Then $|H|=d \Longrightarrow b^d=e$ and $b \in H \Longrightarrow b \in G$. Hence $b=a^k$ for some $k \in \mathbb{Z}$. Also $e=b^d=\left(a^k\right)^d=a^{k d}$ and $|a|=n$ implies $n \mid k d$. Hence $k d=n k^{\prime}$ for some $k^{\prime} \in \mathbb{Z}$ and $k=\left(\frac{n}{d}\right) k^{\prime}$ Hence $b=a^k=a^{\left(\frac{n}{d}\right) k^{\prime}}$ for some $k^{\prime} \in \mathbb{Z}$ implies $b \in H_d$. Hence $H \subseteq H_d$ and then $H=H_d$. $\square$ Corollary: Now number of elements of order $d$ are precisely the number of generators of $H_d$. By [[Order of Power of Group Element]], $a^r \in H_d$ generates $H_d$ iff $1 \leq r \leq d$ and $\operatorname{gcd}(r, d)=1$ $\left(\left|a^r\right|=\frac{d}{\operatorname{gcd}(r, d)}\right.$. Then $\left|a^r\right|=d$ iff $\left.\operatorname{gcd}(r, d)=1\right)$ Hence the number of generators of $H_d$ is exactly $\varphi(d)$ which are precisely the elements of order $d$.