Is a solvable group satisfying a semigroup law?

Question

Let $S$ be the free semigroup on the set $\{x_1,\ldots ,x_n\}$, where $n$ is a positive integer. Suppose that $\mu=\mu (x_1,\ldots ,x_n)$ and $\nu = \nu (x_1,\ldots ,x_n)$ are two elements in $S$. We say that $\mu=\nu$ is a semigroup law in a group $G$ if for every $n$-tuple $(g_1,\ldots , g_n)$ of elements of $G$, we have $\mu (g_1,\ldots , g_n)=\nu (g_1,\ldots , g_n)$.

For example an Abelian group satisfies the semigroup law $\mu=\nu$ where $\mu (x_1,x_2)=x_1x_2$ and $\nu (x_1,x_2)=x_2x_1$. And similarly for nilpotent groups of class $c$ we can find a finite sequence of semigroup laws $\mu_1=\nu_1,\ldots,\mu_c=\nu_c$.

Now I am looking for semigroup laws that solvable groups are expressible by them.

Answer 1

The generators of a free metabelian group of rank two generate a free subsemigroup so there is no semigroup law defining solvable of derived length 2. Much smaller metabelian groups, like the lamplighter group, also have free subsemigroups of rank 2.

Update. It is proved in J. A. Lewin and T. Lewin,' Semigroup laws in varieties of solvable groups', Proc. Cambridge Philos. 5oc. 65(1969), 1-9 that a finitely generated solvable group satisfies a semigroup law iff it is virtually nilpotent so the answer is no in an extreme sense.

Update 2. As pointed out by @YCor, Rosenblatt proved a finitely generated solvable group that is not virtually nilpotent (or in the paper he works with polynomial growth) has a free subsemigroup on two generators. The reference can be found in the comments.