Is there an $n$ such that $D_n$ contains a subgroup isomorphic to $Q_8$?

My immediate thought is no, but I'm not sure how to prove it. I know that there are only $2$ non-Abelian groups of order $8$ (up to isomorphism): $D_4$ and $Q_8$. I feel like the answer should fall out from here but I'm stuck.

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    $\begingroup$ No, $Q_8$ has three cyclic subgroups of order $4$, but there is no $D_n$ with this property: in any $D_n$, the elements outside of $\langle r \rangle$ (the rotation subgroup) all have order $2$, and since $\langle r \rangle$ is cyclic, it contains exactly one subgroup of each possible order. So $D_n$ contains at most one cyclic subgroup of order $4$. $\endgroup$ – Bungo Nov 2 '17 at 0:06
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    $\begingroup$ There is also the more general result stating that every subgroup of a dihedral group is cyclic or dihedral. $\endgroup$ – Gregoire Rad Nov 2 '17 at 0:43

By Theorem 3.1 of K. Conrad's notes, every subgroup of a dihedral group is either cyclic or itself dihedral.

Since $Q_8$ is neither cyclic nor dihedral, it cannot be isomorphic to a subgroup of a dihedral group.

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