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Since the word problem for H is unsolvable, the problem of deciding membership in LH is recursively unsolvable. This proves the theorem. 3 (Mihailova [75]). Let F be a finitely generated free group of rank at least 2. Then the group G = F x F has a finitely generated subgroup L such that the generalized word problem for L in G is recursively unsolvable. The above lemma can also be used to obtain a number of other unsolvability results concerning the direct product of two free groups. 4 (Miller [77]).

If G is a finite extension of the finitely generated group H having solvable generalized word problem, then G has solvable generalized word problem. 11 ([30]). (1) There is a finitely presented group G 1 with unsolvable conjugacy problem that has a subgroup M of index 2 which has solvable conjugacy problem. (2) There is a finitely presented group G 2 with solvable conjugacy problem that has a subgroup L of index 2 which has unsolvable conjugacy problem. An example of the first type was given by Gorjaga and Kirkinskii [41], while examples of both of these phenomena were given by Collins and Miller [30].

As above, either w appears on the list of words equal to 1 in H or else since S is simple U appears on the list of words equal to 1 in Hw (in which case w i-H 1 ). So by enumerating these two lists we can decide whether w is equal to 1 in G. For the converse, suppose G has solvable word problem. Then the set of all pairs of words (u, v) such that u i-e 1 and v i-e 1 is recursive and can be arranged in a recursive list as say (Ui,Vi),i = 1,2, .... Let X,tl,t2, ... be new generating symbols and form the presentation u(G) =< G,x,tl,t2,··.