# Show that an element in $G'$ cannot be expressed as a single commutator $[x,y]$.

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 $$x,\,y\in G$$.

What I am confused is that how to check that $$i^2a$$ cannot be expressed as a single commutator $$[x,y]$$, for any $$x,\,y\in G$$? The annotation in the book says this is an elementary calculation. I think it may be prove by contradiction.

Suppose $$i^2a=(-1,a,1,1)=[(h_1,k_1),\,(h_2,k_2)]$$. After a tedious calculation, I obtained $$(-1,a,1)=(k_1^{-1}\cdot h_1^{-1})\lbrack(k_1^{-1}k_2^{-1})\cdot(h_2^{-1}h_1)\rbrack(k_2^{-1}\cdot h_2)$$. There are too many possibilities, so how to check this assertion briefly?

• Note that $C_3$ acts on the $Q_8$ and the $V$ factors separately. So you can separate out $Q_8\rtimes C_3$ and $V\rtimes C_3$, except you need to make sure the $C_3$ stays the same to patch them back together. – user10354138 May 17 at 9:06
• What is the $V$? – Tao X May 18 at 13:59
• – user10354138 May 18 at 14:01
• Can you give an answer? I still don't make any progress. – Tao X May 19 at 8:40