TITLE: Conjugacy problem in a conjugacy separable group
QUESTION [20 upvotes]: Here is a question that has been bothering me for some time:
Let G be a finitely generated conjugacy separable group with solvable word problem. Does it follow that the conjugacy problem in G is solvable?
Background. A group G is said to be conjugacy separable is for any two non-conjugate elements x,y in G there is a homomorphism from G to a finite group F such that the images of x and y are not conjugate in F. Equivalently, G is conjugacy separable if each conjugacy class is closed in the profinite topology on G.
A well-known theorem of Mal'cev states that a finitely presented conjugacy separable group has solvable conjugacy problem (in this case it is possible to recursively enumerate all the finite quotients, simultaneously checking conjugacy of the images of two given elements in each of them).
On the first page of the paper 'Conjugacy separability of certain torsion groups.'
(Arch. Math. (Basel) 68 (1997), no. 6, 441--449.) Wilson and Zalesskii claim that the conjugacy problem is solvable in finitely generated recursively presented conjugacy separable groups (which, of course, implies a positive answer to my question), and refer to a work of J. McKinsey, 'The decision problem for some classes of sentences without quantifiers' (J. Symbolic Logic 8, 61 – 76 (1943)). 
However, I could not find anything in the latter paper that would allow to deal with infinite recursive presentations. Moreover, the corresponding property for residually finite groups simply fails. More precisely, there exist finitely generated residually finite recursively (infinitely!) presented groups with unsolvable word problem (cf. 'A Finitely Generated Residually Finite Group with an Unsolvable Word Problem' by S. Meskin, Proceedings of the American Mathematical Society, Vol. 43, No. 1 (Mar., 1974), pp. 8-10).

REPLY [3 votes]: This is another incomplete answer, but which hopefully can still be useful.
Say that a finitely generated group G has computable finite quotients if there is an algorithm that, on input a finite group F given by a finite presentation and a function that maps the generators of G to those of F, decides whether or not this function could be extended as a homomorphism.
McKinsey’s argument then clearly yields: in a finitely generated, conjugacy separable group, with computable finite quotients, there is an algorithm that decides when two elements are not conjugated.
And thus if that group is recursively presented, it must have solvable conjugacy problem.
I studied this property in my article: « Computability of finite quotients of finitely generated groups », where I prove that this property is independent from the solvability of the word problem: a residually finite group can have computable finite quotients while having unsolvable word problem, and there exist finitely generated residually finite groups with solvable word problem but without computable finite quotients. (This answers negatively Ian Agol’s question.)
Because of this, we have no reasons to believe that for finitely generated groups with solvable word problem, being conjugacy separable is a sufficient condition to have solvable conjugacy problem, since this hypothesis cannot be used without the ability to detect finite quotients.
However, no matter how much I would like to plainly answer « no» to your question, I don’t know how to change my construction to build a conjugacy separable group with solvable word problem and without computable finite quotients, less again one with unsolvable conjugacy problem.
Note finally that it follows from an article of René Hartung (Coset enumeration for certain infinitely presented groups, International Journal of Algebra and Computation) that the first Grigorchuk group has computable finite quotients, and thus the article of Wilson and Zalesskii which you mention, and which proves that this group is conjugacy separable, does help proving that it has solvable conjugacy problem.