Let's define the groupof quaternions $Q_8$: $Q_8$ is the group generated by the 2 matrices $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 & \end{bmatrix}$ and $B = \begin{bmatrix} 0 & i \\ i & 0 & \end{bmatrix}$

I have to show that all the subgroups of $Q_8$ are cyclic.

Here's my attempt:

First let's make a list of all the elements of $Q_8$.

$A^4 = id$

$B^2 = A^2 \implies B^4 = id$

$A^3B = BA $

$\therefore$ $Q_8 = \{id , A, A^2, A^3, AB, A^2B, BA, B\}$

We therefore have the following subgroups (attempt atfinding the subgroups):

  • $\{id\}$

  • $Q_8$

  • $\{id, B^2\}$

  • $\{id, A^2\}$

  • $\{A, A^3, A^2, id\}$

I probably forgot a bunch of them but is that the right approach to do it, just write out all the subgorups and show that they're each generated by one elemment.

Also, how is $Q_8$ cyclic if it's only generated by one element?

  • 1
    $\begingroup$ $\{id, A, A^3\}$ is not a subgroup because it is not closed under the group operation. For example, if you do $A\cdot A$, you get $A^2$, which is not inside the set. Therefore, the group operation gives you elements outside the set, so the set is not a group. $\endgroup$ – Noble Mushtak Jun 15 '17 at 14:35
  • 1
    $\begingroup$ Also, $Q_8$ itself is not cyclic since it is generated by two elements, not just one. I think what you should actually show is that all strict subgroups (i.e. subgroups other than $Q_8$) are cyclic. $\endgroup$ – Noble Mushtak Jun 15 '17 at 14:37

The quaternion group is given by $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$. By Lagrange, proper subgroups have order $2$ or $4$. Since all groups of order $2$ are cyclic, we only need to look for subgroups of order $4$. So we are left with the group $$ \langle i \rangle= \{\pm 1, \pm i\} $$ which is generated by $i$ or $-i$. By definition, it is cyclic. The same holds for $\langle k\rangle$ and $\langle j\rangle$. Indeed, we have found all subgroups of order $4$, see this duplicate.

Edit: See here, how to rewrite $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$ into matrices as above.


@DietrichBurde's answer is sufficient, but I think it might be helpful to see this in terms of $A$ and $B$ like in the question.

By Lagrange's theorem, all proper subgroups are either of order $1$, $2$, or $4$.

Subgroups of Order $1$: $\{id\}$ -> Obviously cyclic.

Subgroups of Order $2$: $\{id, A^2\}$ -> Again, obviously cyclic.

Note that I did not list $\{id, B^2\}$. This is because $A^2=B^2$, so this is the same group.

Subgroups of Order $4$: $\{id, A, A^2, A^3\}$ and $\{id, B, A^2, A^2B\}$ and $\{id, AB, A^2, BA\}$

The first group is generated by $A$ (or $A^3$), the second by $B$ (or $A^2B$), and the third by $AB$ (or $BA$). Therefore, all of these groups are cyclic.

Now, finally, we need to prove that we did not miss any subgroups.

We have listed all the groups of order $1$, since there can be only one such group, which is the trivial group.

All groups of order $2$ are isomorphic to $\Bbb{Z}_2$, meaning it has one element of order $2$. If we go through all of the elements in $Q_8$, the only one with order $2$ is $A^2$, so $\{id, A^2\}$ is the only subgroup of order $2$.

All groups of order $4$ are isomorphic to $\Bbb{Z}_4$ or $\Bbb{Z}_2 \times \Bbb{Z}_2$.

  • $\Bbb{Z}_4$ is generated by an element of order $4$. There are $6$ such elements in $Q_8$: $A, A^3, AB, BA, B, A^2B$. We have listed all subgroups generated by these elements, so this has been exhausted.
  • $\Bbb{Z}_2 \times \Bbb{Z}_2$ has three different elements of order $2$. However, as stated above, $Q_8$ has only one element of order $2$, so this can't happen. Thus, there are no subgroups in $Q_8$ isomorphic to $\Bbb{Z}_2 \times \Bbb{Z}_2$.

Therefore, we have listed all proper subgroups of $Q_8$ and shown all of them are cyclic, so we are done.


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.