Each Finite Cyclic Group of Order $n$ Contains Unique Subgroup of Order $d$ Where $dmid N$

$bullet$ Indeed, $(g^e)^dg^edg^n1$. From that, we have that $|g^e||H_d|leq d$.$bullet$ Suppose $|g^e|k$. Therefore, $g^ek1$. Since $|g|n$, we have that $nleq ek$. It follows that $dfracnedleq k$. So $|H_d||g^e| kgeq d$.Conclusion: $|H_d|d$.$mathbfUnicity$: Suppose $H$, where $k$ is the smallest positive integer such that $g^kin H$.$bullet$ Since $|H|d$, we get that $g^kd1.$ But $|g|n$ and $g^kd1$, it follows that $kdgeq n$.

$bullet$ By the same way, $d$ is the smallest positive integer such that $(g^k)^dg^kd1$. And since $g^n1$, it follows that $kdleq n$. From that, we get that $kdn$ which implies that $kdfracnde$. Finally, we get that $H_dH$.

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 $supseteq$.

·OTHER ANSWER:

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 $supseteq$.