Let $G$ be a group, let $G^{(1)}:=[G,G]$ be its commutator subgroup and let $G^{(2)} :=[G^{(1)},G^{(1)}]$. I have read in a book that group-theoretic conjugation, understood as an action of $G$ on $G$ induces an action of $G/G^{(1)}$ on $G^{(1)}/G^{(2)}$.
Can someone explain me why is this? If $\bar{x} \in G/G^{(1)}$ and $[aba^{-1}b^{-1}] \in G^{(1)}/G^{(2)}$ then $$ \bar{x} \cdot [aba^{-1}b^{-1}] = [xaba^{-1}b^{-1}x^{-1}] $$ and I don't even see this is an element of $G^{(1)}$.
The book in question is Lickorish' An Introduction to Knot Theory, page 116.
The (left) conjugation action ${}^gx:=gxg^{-1}$ intertwines the commutator, i.e. $ {}^g[a,b]=[{}^ga,{}^gb]$ (and same with the right conjugation action, regardless of which definition for $[a,b]$ used).
Indeed, if $w(a,b,\cdots)$ is any word function (product of integer powers of the arguments) and $\phi$ is any homomorphism (with the same domain as $w$'s arguments), then
$$\phi(w(a,b,\cdots))=w(\phi(a),\phi(b),\cdots)$$ Good exercise. This applies of course with $\phi_g(x)={}^gx$ and $w(a,b)=[a,b]=aba^{-1}b^{-1}$.