So I have finitely presented group with 2 generators. Can I solve word problem in it (check if two words X and Y are actually the same element of my group)?
It is a classical result, originally by HigmanNeumannNeumann, that every finitely presented group embeds in a 2generator finitely presented group. In particular, every finitely presented group with unsolvable word problem embeds in a 2generator group. Given that having solvable word problem is preserved by subgroups, such 2generator groups must also have unsolvable word problem.
Moreover, this construction is uniform: given a finite presentation P, we can construct a new 2generator finite presentation Q into which P embeds (as well as an explicit embedding). Also, Q has m+n relators, where m is the number of generators of P, and n is the number of relators of P. However, the relators of Q can get very long.
You can find their original paper at: G. Higman, B. H. Neumann, H. Neumann, "Embedding theorems for groups", J. London Math. Soc. 24, 247254 (1949).
Maurice
In general the answer is no.
A counterexample can be found in the paper by W. Boone "The word problem" (Ann. of Math. (2) 70 (1959) 207–265).
Quoting from page 210:
"It thus follows from Result c (using the embedding result of [8] noted above) that one can exhibit a finite presentation of a group consisting of two generators and thirtytwo defining relations and having an unsolvable word problem."

$\begingroup$ I need only general answer currently. So thank you! $\endgroup$ – ptashek Jul 22 '11 at 11:15