# Is a solvable group satisfying a semigroup law?

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.

• 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.
– YCor
Oct 7, 2021 at 9:01
• Thanks for your comment, I changed it @YCor Oct 7, 2021 at 9:57
• 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.
– YCor
Oct 7, 2021 at 10:06

• 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