If $g_i$ generate $G$, then $x^kg_i^\pm x^{-k}$ generate $[G,G]$

Question

Suppose that under the quotient map $q: G\to [G,G]$, $g\mapsto [g]$, we have an element $x\in G$ such that $q(x)$ is a generator of $[G,G]$. Then we are asked to show that elements of the form $x^kg_i^\pm x^{-k}$ generate $[G,G]$ (here $g_i$ are generators of $G$). I am not really sure how to advance on this. I tried writing $$ x^k g_i x^{-k}=x^k g_i x^{-k}g_i^{-1}g_i=[x^k, g_i] g_i $$ but I am not really sure how to proceed from here. I would appreciate a hint. Thanks!

Answer 1

This will be difficult to prove because it's not true in general. Consider, for example, the smallest non-abelian group, the group $S_3$ of permutations of a $3$-element set. Its commutator subgroup is generated by either of the two $3$-cycles, so take one of those as $x$. The group $S_3$ itself is generated by any two of the three $2$-cycles, so take two of those as your $g_i$'s. Then your alleged generators $x^kg_i^\pm x^{-k}$ of $[S_3,S_3]$, being a conjugates of a $2$-cycle, are themselves $2$-cycles and are therefore not in $[S_3,S_3]$.

In general, the alleged generators $x^kg_i^\pm x^{-k}$ of $[G,G]$ will actually generate all of $G$. So the claimed result fails unless $[G,G]=G$.