Proof
(related to Proposition: Subgroups of Finite Cyclic Groups)
 Let \((G,\ast)\) be a finite cyclic group generated by \(g\), i.e. \(G=\langle g \rangle\) with \(G=n\).
 Furthermore, let \(d\mid n\) be a divisor of $n$, there is a \(k\in\mathbb Z\) with \(d\cdot k=n\).
Existence
 Following the lemma about subgroups of cyclic groups, the element \(g^k\in G\) generates a subgroup \(H=\langle g^k\rangle \subseteq G\).
 Because \((g^k)^d=g^{kd}=g^n=e\), the order of \(g^k\) in \(G\) is \(d\).
 Thus, \(H=d\).
Uniqueness
 If there was another subgroup \(H'\subseteq G\) with \(H'=d\), but \(H'\neq H\), then it would be generated by the element \(g^j\in G\) with \(j\neq k\).
 This is a contradiction to \(dj=n=dk\), which is equivalent to \(j=k\).
∎
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013