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.

  • $\begingroup$ You write "expressible by a semigroup law" but this is confusing, it should be "satisfying a semigroup law". Indeed the law is far from characterizing the group, as the verb "express" suggests. $\endgroup$
    – YCor
    Oct 7, 2021 at 9:01
  • $\begingroup$ Thanks for your comment, I changed it @YCor $\endgroup$ Oct 7, 2021 at 9:57
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    $\begingroup$ Note also that "expressible by a set $S$ of semigroup law" could make sense, but not for a group, rather for a class $C$ of groups, namely $C$ would be exactly the class of groups satisfying all laws in $S$. For instance, abelian (=1-step-nilpotent) groups are those satisfying $xy=yx$, and $c$-step-nilpotent groups (=those of class $\le c$) are those satisfying a certain finite set of semigroup laws, if I remember correctly. $\endgroup$
    – YCor
    Oct 7, 2021 at 10:06

1 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.

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    $\begingroup$ Rosenblatt (1974) even improved the latter fact, showing that any non-virtually-nilpotent f.g. solvable group contains a free subsemigroup on 2 elements (and hence on $k$ elements for any $k$). DOI link pointing to AMS site $\endgroup$
    – YCor
    Oct 6, 2021 at 16:00
  • $\begingroup$ @YCor, I had forgotten that. Thanks $\endgroup$ Oct 6, 2021 at 16:24

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