1
$\begingroup$

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)?

$\endgroup$
0
$\begingroup$

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).

-Maurice

$\endgroup$
11
$\begingroup$

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."

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.