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Homesubgroups of finite cyclic group

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# subgroups of finite cyclic group

Let $n$ be the order of a finite cyclic group $G$. For every positive divisor $m$ of $n$, there exists one and only one subgroup of order $m$ of $G$. The group $G$ has no other subgroups.

*Proof.* If $g$ is a generator of $G$ and $n=mk$, then $g^{k}$ generates the subgroup $\langle g^{k}\rangle$, the order of which is equal to the order of $g^{k}$, i.e. equal to $m$. Any subgroup $H$ of $G$ is cyclic (see this entry). If $|H|=m$, then $H$ must have a generator of order $m$; thus apparently $H=\langle g^{{\pm k}}\rangle=\langle g^{k}\rangle$.

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## Mathematics Subject Classification

20A05*no label found*

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