0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user10354138 May 17 at 9:06
  • $\begingroup$ What is the $V$? $\endgroup$ – Tao X May 18 at 13:59
  • $\begingroup$ The Klein Vierergruppe. $\endgroup$ – user10354138 May 18 at 14:01
  • $\begingroup$ Can you give an answer? I still don't make any progress. $\endgroup$ – Tao X May 19 at 8:40

Your Answer

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

Browse other questions tagged or ask your own question.