We prove that a countable group G can be embedded into a two-generator simple group S which is an amalgamated free product of groups G * F and F, where F and F are free groups on two generators.S is also the product of two commuting free subgroups.If G has solvable word problem, then we can construct a recursive presentation for S.1. Introduction.It is well known that every countable group is a subgroup of a simple two-generator group (see [3] and [7]).The main result of our paper is Theorem A. (a) A countable group G can be embedded into a simple two-generator group S which is an amalgamated free product of G * Fx and F, where F and Fx are both free groups on two generators.S is also the product of two commuting free subgroups.(b) If G has solvable word problem, we can construct the group S of (a) with a recursive presentation.It was first shown by R. J. Thompson [8] by a construction completely different from ours that a countable group with solvable word problem is a subgroup of a recursively presented finitely generated simple group.Our construction of the group S is similar to the construction of the simple free products of R. Camm [1], or see [6] for a slightly simplified version.2. The construction of the group S. Let 77 be a countable infinite group with an element x of order at least 5, and assume that the elements of 77 are numbered by the integers, 77 = {g" i G Z}, such that g0 = 1, gx = x, g2 = x2, g3 = x3, g4 = x4.Let A = (a) and F = (b,c) be free groups on a and b, c, respectively.Then the subgroup U of A * H generated by the elements of the form ", = aW *'eZ-{0}, where p is a permutation of Z -{0}, is a free group freely generated by the elements u" / g Z -{0}, with transversal AinA*H.Similarly, the elements Vj = bjcu\ 7GZ-{0), of F, where a is a second permutation of Z -{0}, freely generate a free subgroup V of F with transversal B = (b).