Radicals and Plotkin's problem concerning geometrically equivalent groups (Q2759012)

If \(G\) and \(X\) are groups and \(N\) is a normal subgroup of \(X\), then the \(G\)-closure of \(N\) in \(X\) is the normal subgroup \(\overline X^G=\bigcap\{\ker\varphi\mid\varphi\colon X\to G,\;N\subseteq\ker\varphi\}\). In partcular, \(\overline 1^G=R_GX\) is the \(G\)-radical of \(X\). B. Plotkin called two groups \(G\) and \(H\) geometrically equivalent, written \(G\sim H\), if for any free group \(F\) of finite rank and any normal subgroup \(N\) of \(F\) the \(G\)-closure and the \(H\)-closure of \(N\) in \(F\) are the same, i.e., \(\overline N^G=\overline N^H\). It is easy to see that \(G\sim H\) if and only if \(R_GK=R_HK\) for all finitely generated groups \(K\). B. Plotkin noted that geometrically equivalent groups satisfy the same quasi-identities and showed that nilpotent groups \(G\) and \(H\) satisfy the same quasi-identities if and only if \(G\) and \(H\) are geometrically equivalent. This led B. Plotkin to conjecture that this might hold for any pair of groups [see Problem 14.17 in the Kourovka Notebook, Russian Academy of Science, Novosibirsk, 14th ed. (1999; Zbl 0943.20003)]. The authors refute this conjecture. Namely they prove the exictence of groups \(G\), \(L\), \(H=L\times G\) such that \(R_GL=L\) and \(R_HL=1\) (in particular \(G\) and \(H\) are not geometrically equivalent) and \(G\) and \(H\) satisfy the same quasi-identities.NEWLINENEWLINENEWLINEReviewer's remark: The problem of geometrically equivalent algebraic systems (in particular groups) is completely solved by \textit{A. Myasnikov} and \textit{V. Remeslennikov} [J. Algebra 234, No. 1, 225-276 (2000; Zbl 0970.20017)] (see Theorems C1, C2, and C4). In the same place, concrete examples of groups, refuting the conjecture of B. Plotkin, are constructed.