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 Higman-Neumann-Neumann, that every finitely presented group embeds in a 2-generator finitely presented group. In particular, every finitely presented group with unsolvable word problem embeds in a 2-generator group. Given that having solvable word problem is preserved by subgroups, such 2-generator groups must also have unsolvable word problem.

Moreover, this construction is uniform: given a finite presentation P, we can construct a new 2-generator 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, 247-254 (1949).



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 thirty-two 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

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