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> [!proposition] Proposition. ([[uniqueness of subgroups of finite cyclic groups]])
> Let $G$ be a [[cyclic group]] of [[order of a group|order]] $n$. Then every [[subgroup]] of $G$ is [[cyclic group|cyclic]]. Additionally, if $d | n$, then $G$ has a *unique* (cyclic) [[subgroup]] of order $d$.
^90b418
> [!proof]- Proof. ([[uniqueness of subgroups of finite cyclic groups]])
> Every [[subgroup]] of $C_{n}$ is [[cyclic group|cyclic]]. Moreover, for any $d | n$, $G$ has a *unique* ([[cyclic group|cyclic]]) [[subgroup]] of [[order of a group|order]] $d$.
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Let $H \leq G$ have order $m$. Choose $a^{i} \in H$ such that $|a^{i}| \geq |a^{j}|$ for all $a^{j} \in H$. Since $H$ is closed under the [[binary operation|group law]], we have that the cyclic subgroup $\{a^{i}, a^{2i},\dots,a^{ni / \text{gcd}(i,n)}=e)\} \leq H$. By [[Lagrange's Theorem]], $|a^{i}|=k|a^{j}|$ for some $k \in \mathbb{Z}$. Hence $\textcolor{Apricot}{a^{j}=(a^{i})^{k}}$. Because $a^{j}$ was arbitrary, we conclude that $H \geq \{a^{i}, a^{2i},\dots,a^{ni / \text{gcd}(i,n)}=e)\}$ as well. Thus the two-way inclusion is shown— $H$ is the [[cyclic group]] [[generating set of a group|generated by]] $a^{i}$.
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Let $d$ be such that $d | n$. Write $n=kd$. Any [[subgroup]] $H_{d}$ of order $d$ needs to have $d$ elements and be generated by one of them.
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$H_{d}=\{ s,s^{2},\dots,s ^{d}=a^{dk}=e \}.$
It follows that the $s_{\iota}$ are precisely the multiples of $k$, $H_{d}=\{ a^{k}, a^{2k}, \dots, a^{kd}=e \}$where $k$ is such that $n=kd$. Since $H_{d}$ was initially arbitrary, we conclude that this the unique [[subgroup]] of order $d$.
^922997
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```