I was reading the section in Abstract Algebra by Dummit and Foote about semidirect products (section 5.5), and they give the following example of a non-abelian group in which the subgroup generated by the commutators is not the same as the set of commutators (page 180 in the 3rd edition).
Let $H=Q_8\times(Z_2\times Z_2)=\langle i,j\rangle\times(\langle a\rangle\times\langle b\rangle)$ and let $K=\langle y\rangle\cong Z_3$. The map defined by $$ i\mapsto j\qquad j\mapsto k=ij\qquad a\mapsto b\qquad b\mapsto ab $$ is easily seen to give an automorphism of $H$ of order $3$. Let $\varphi$ be the homomorphism from $K$ to $\text{Aut}(H)$ defined by mapping $y$ to this automorphism, and let $G$ be the associated semidirect product, so that $y\in G$ acts by $$ y\cdot i=j\qquad y\cdot j=k\qquad y\cdot a=b\qquad y\cdot b=ab. $$ The group $G=H\rtimes K$ is a non-abelian group of order $96$ with the property that the element $i^2a\in G'$ but $i^2a$ cannot be expressed as a single commutator $[x,y]$, for any $x,\,y\in G$ (checking the latter assertion is an elementary calculation).
What I am confused about is how to check that $i^2a$ cannot be expressed as a single commutator $[x,y]$, for any $x,\,y\in G$? Dummit and Foote claim that this is an elementary calculation, but I'm struggling with it.
My thought process goes as follows. I think it can be proved by contradiction.
Suppose $i^2a=[(h_1,k_1),\,(h_2,k_2)]$ where $h_1,h_2 \in H, k_1, k_2 \in K$. After a tedious calculation, I obtained $i^2a=(k_1^{-1}\cdot h_1^{-1})((k_1^{-1}k_2^{-1})\cdot h_2^{-1})(k_1^{-1}k_2^{-1}) \cdot (h_1 k_2 \cdot h_2)$.
I'm not quite sure where to go from here. I don't see an obvious line of attack here, and it doesn't seem like what I've done so far will turn out to truly be an elementary calculation as Dummit and Foote state.
Does anybody have any thoughts to share please? Any insights would be much appreciated.