TITLE: Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$
QUESTION [5 upvotes]: A cyclic group $G := \langle x \rangle$ of finite order $n$ has a
  unique subgroup of order $d$, namely $$\langle x^{n/d} \rangle = \{g
 \in G : g^d = 1\}$$ for every divisor $d$ of $n$.

I wanted to show the equality $\langle x^{n/d} \rangle = \{g \in G : g^d = 1\}$. Now for the inclusion $\subseteq $ we take an element of $\langle x^{n/d} \rangle$ which has the form $x^{kn/d}$ for some natural number $k$. Hence $$(x^{kn/d})^d = x^{kn} = (x^n)^k =1^k = 1$$ and so $x^{kn/d} \in \{g \in G : g^d = 1\}$. However I am a bit lost proving the inclusion $\supseteq$.

REPLY [5 votes]: Assume $G$ is a finite cyclic group and $|G|=n$. There exists $g\in G$ such that $G=<g>$ and $|g|=n$. 
$\mathbf{Existence}$: Put $e=\dfrac{n}{d}$ and define $H_d=<g^e>$. We will show that $|H_d|=d$.
$\bullet$ Indeed, $(g^e)^d=g^{ed}=g^n=1$. From that, we have that $|g^e|=|H_d|\leq d$.
$\bullet$ Suppose $|g^e|=k$. Therefore, $g^{ek}=1$. Since $|g|=n$, we have that $n\leq ek$. It follows that $\dfrac{n}{e}=d\leq k$. So $|H_d|=|g^e|= k\geq d$.
Conclusion: $|H_d|=d$.
$\mathbf{Unicity}$: Suppose $H<G$ and $|H|=d$. We will show that $H=H_d$, where $H_d$ is defined above. We know that $H$ is cyclic. Therefore, $H=<g^k>$, where $k$ is the smallest positive integer such that $g^k\in H$.
$\bullet$ Since $|H|=d$, we get that $g^{kd}=1.$ But $|g|=n$ and $g^{kd}=1$, it follows that $kd\geq n$.
$\bullet$ By the same way, $d$ is the smallest positive integer such that $(g^k)^d=g^{kd}=1$. And since $g^n=1$, it follows that $kd\leq n$. 
From that, we get that $kd=n$ which implies that $k=\dfrac{n}{d}=e$. Finally, we get that $H_d=H$.