Subgroup of $D_n$ isomorphic to $Q_8$.

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.

• 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 $4$. – Bungo Nov 2 '17 at 0:06
• There is also the more general result stating that every subgroup of a dihedral group is cyclic or dihedral. – Gregoire Rad Nov 2 '17 at 0:43

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