Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$ 
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 $\supseteq$.

Assume $g$ satisfies $g^d = 1$. Since $G$ is cyclic, you can find an integer $m$ so that $g = x^m$. But this means $x^{md} = 1$ and since $n$ is the order of $x$, this implies $n | md$. We can now conclude $g = x^m = x^{(n/d) \cdot (dm/n)} \in \langle x^{n/d}\rangle$.
