4
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.