An inverse semigroup is an algebra with two operations: binary $\cdot$ and unary $^{-1}$ such that $\cdot$ is associative and $xx^{-1}x=x, xx^{-1}yy^{-1}=yy^{-1}xx^{-1}$. The Brandt semigroup with 1, $B_2^1$, is the inverse semigroup of $2\times 2$-matrices consisting of 0, I, and the four matrix units $e_{i,j}$, $i,j=1,2$ where $e_{i,j}$ is the matrix with $(i,j)$-entry 1 and other entries 0, $e_{i,j}^{-1}=e_{j,i}$. It is known (Kleiman) that the identities of $B_2^1$ are not finitely based.

Question. Is it known that the identities of any finite inverse semigroup containing $B_2^1$ as an inverse subsemigroup are not finitely based?


I believe this is a well known open question. It has this property as a semigroup but it is not clear as an inverse semigroup. Mark Sapir showed it is contained in a finitely based locally finite variety of inverse semigroups. Your question is problem 3.10.13 in his book Combinatorial algebra: syntax and semantics.

| cite | improve this answer | |
  • $\begingroup$ Thank you! The book is 6 years old. Is it true that the strongest result in this area is by Kadourek about finite inverse semigroups with solvable subgroups? $\endgroup$ – user158834 Jun 9 at 15:37
  • $\begingroup$ I'm not aware of anything more recent but I don't really follow this very closely. You might email Sapir. He probably follows more closely. This problem has been open for a long long time so I doubt much has changed. $\endgroup$ – Benjamin Steinberg Jun 9 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy