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I tried the dihedral group $D_3$ of order $6$ which is not cyclic. I take the subgroup containing the three rotations ($R_0$, $R_{120}$, $R_{240}$) it is cyclic. Does it work for all groups?

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closed as unclear what you're asking by Dietrich Burde, Claude Leibovici, Saad, Shailesh, Leucippus Mar 10 '18 at 0:39

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    $\begingroup$ Yes, any group of order $p$ is cyclic, in particular such subgroups. For dihedral groups $D_n$, the subgroup of rotations $C_n$ is always cyclic, not only for $n=3$. $\endgroup$ – Dietrich Burde Mar 9 '18 at 13:27
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    $\begingroup$ Every group of prime order is cyclic. A subgroup of prime order is a group. Hence it is cyclic. $\endgroup$ – Levent Mar 9 '18 at 13:28
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A subgroup is itself a group. The "sub-" prefix is just there to indicate that the underlying set is a subset of a larger set.

Since the theorem works for any group and a subgroup is again a group, the subgroup has to behave according to the theorem regardless of the larger group that the subgroup is contained.

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  • $\begingroup$ What is the reason for the downvote ? $\endgroup$ – onurcanbektas Mar 9 '18 at 14:06
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Consider a group $G$ of order $n$, where $n$ is not prime and $G$ is non-cyclic. Let $N$ be any subgroup of $G$ of prime order $p$. By Lagrange's Theorem, there can only be two subgroups of $N$, $N$ itself and $\{e\}$. It thus follows that there exists an element $x \in N$ which generates $N$. Thus $N$ is a cyclic group of prime order, i.e. $N \simeq C_p$.

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