14.71 (1999)

Consider a free group $F$ of finite rank and an arbitrary group $G$. Define the $G$-closure $\text{cl}_G(T)$ of any subset $T \subseteq F$ as the intersection of the kernels of all those homomorphisms $\mu : F \to G$ of $F$ into $G$ that vanish on $T$: $$\text{cl}_G(T) = \bigcap \{ \text{Ker } \mu \mid \mu : F \to G; T \subseteq \text{Ker } \mu \}.$$ Groups $G$ and $H$ are called geometrically equivalent if for every free group $F$ and every subset $T \subset F$ the $G$- and $H$-closures of $T$ coincide: $\text{cl}_G(T) = \text{cl}_H(T)$. It is easy to see that if $G$ and $H$ are geometrically equivalent then they have the same quasiidentities. Is it true that if two groups have the same quasiidentities then they are geometrically equivalent? This is true for nilpotent groups.