19.64 (2018)
OpenLet $G$ be a group, and $(g_1, \dots, g_n)$ a tuple of its elements. The type of this tuple in $G$, denoted $Tp^G(g_1, \dots, g_n)$, is the set of all first order formulas in free variables $x_1, \dots, x_n$ in the standard group theory language which are true on $(g_1, \dots, g_n)$ in $G$. Two groups $G$ and $H$ are called isotypic if for every tuple of elements $\bar{h} = (h_1, \dots, h_n)$ in $H$ there is a tuple $\bar{g} = (g_1, \dots, g_n)$ in $G$, such that $Tp(\bar{h}) = Tp(\bar{g})$ and vice versa, for every tuple $\bar{g}$ in $G$ there is a tuple $\bar{h}$ in $H$ such that $Tp(\bar{h}) = Tp(\bar{g})$. Is it true that every two isotypic finitely generated groups are isomorphic?
Progress
The answer is positive if one of the finitely generated groups is free (R. Sklinos), abelian (G. Zhitomirski), virtually polycyclic, metabelian, free solvable (A. Myasnikov, N. Romanovskii), co-Hopfian (in particular homogeneous), finitely presented Hopfian or geometrically Noetherian Hopfian (R. Sklinos). In particular, torsion-free hyperbolic groups, braid groups, linear groups, mapping class groups of compact surfaces, limit groups, quasi-cyclic groups, $\text{Out}(F_n)$, $n > 2$, irreducible arithmetic lattices (which are not virtually free groups) in semi-simple Lie groups, are of such kind (R. Sklinos).
Proof claims
No proof claims yet.
Log in to claim a proof.
Comments
No comments yet. Be the first to comment.
Log in to post a comment.