2.16 (1966)

Solved

A group $G$ is called congruacy separable if any two of its elements are conjugate in $G$ if and only if their images are conjugate in every finite homomorphic image of $G$. Is $G$ conjugacy separable in the following cases:
$\qquad$ a) $G$ is a polycyclic group,
$\qquad$ b) $G$ is a free soluble group,
$\qquad$ c) $G$ is a group of (all) integral matrices,
$\qquad$ d) $G$ is a finitely generated group of matrices,
$\qquad$ e) $G$ is a finitely generated metabelian group?

Progress

a) Yes (V. N. Remeslennikov, Algebra and Logic, 8 (1969), 404–411; E. Formanek, J. Algebra, 42 (1976), 1–10).

b) Yes (V. N. Remeslennikov, V. G. Sokolov, Algebra and Logic, 9 (1970), 342–349).

c), d) Not always (V. P. Platonov, G. V. Matveev, Dokl. Akad. Nauk BSSR, 14 (1970), 777–779 (Russian); V. N. Remeslennikov, V. G. Sokolov, Algebra and Logic, 9 (1970), 342–349; V. N. Remeslennikov, Siberian Math. J., 12 (1971), 783–792).

e) Not always. Let $p$ be a prime and let $A_1, A_2$ be two copies of the additive group $\{m/p^k \mid m, k \in \mathbb{Z}\}$. Let $b_1, b_2$ be the automorphisms of the direct sum $A = A_1 \oplus A_2$ defined by $a^{b_2} = pa$ for any $a \in A$ and $a_2^{b_1} = a_1 + a_2$ and $a_1^{b_1} = a_1$ for some fixed elements $a_1 \in A_1, a_2 \in A_2$. Let $G$ be the semidirect product of $A$ and the direct product $\langle b_1 \rangle \times \langle b_2 \rangle$ (of two infinite cyclics). It can be shown that the elements $a_2$ and $a_2 + a_1/p$ are not conjugate in $G$, but their images are conjugate in any finite quotient of $G$. (M. I. Kargapolov, E. I. Timoshenko, Abstracts of the 4th All-Union Sympos. Group Theory, Akademgorodok, 1973, Novosibirsk, 1973, 86–88 (Russian).)

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