How to find the derived subgroup of $Q_8$

Question

Given the group $Q_8=\{ \pm1, \pm i, \pm j, \pm k \}$ is there an easy way to find its derived subgroup?

I've tried by hand and this seems such a big task to do manually.

After this, by thinking a little bit I've gotten to the point where I've realized that if I got a subgroup of low enough order where its quotient group is abelian I could narrow the search for the derived group but still I fail to do so.

Thanks for the help in advance

Answer 1

The center $Z = \{\pm 1\}$ has order $2$, so the quotient group $Q_8/Z$ has order $4$ and thus is abelian (whether it's cyclic or not doesn't matter!), so all commutators in $Q_8$ become trivial in $Q_8/Z$. Thus the derived subgroup $Q_8'$ has to be inside $Z$. Since $$ [i,j] = iji^{-1}j^{-1} = ij(-i)(-j) = ijij = kk = -1, $$ we have $Z = \{\pm 1\} \subset Q_8'$. Hence $Q_8' = Z = \{\pm 1\}$.

Answer 2

First, since $Q_8$ is not abelian, the derived subgroup must be non-trivial. Since $Q_8$ has non-trivial abelian factor groups, the derived subgroup can't be all of $Q_8$. Moreover, it's easy to see by brute force that $-1$ is an element of the derived subgroup.

Note that $\sigma:Q_8 \to Q_8$ defined by $\sigma(i)=j, \sigma(j)=k, \sigma(k)=i, \sigma(-1)=-1$ is an automorphism of $Q_8$. If $i \in [Q_8, Q_8]$, then so is $j=\sigma(i)$. But then $\langle -1, i, j \rangle= Q_8 \subseteq [Q_8, Q_8]$, which we just determined is impossible. Thus, $i \notin [Q_8, Q_8]$. Similarly, $j, k \notin [Q_8, Q_8]$, so $[Q_8, Q_8] = \{ 1, -1 \}$.