Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear All!

At the time when Lyndon and Schupp wrote their book there was an open question:

Question: Does every finitely presented group with soluble word problem embed in a finitely presented simple group?

Is it still open? Could you hint at some useful references about this? Thanks!

share|improve this question
1  
I assume from the phrasing of your question that there is known a finitely presented group (without soluble word problem) that does not embed in a finitely presented simple group? –  Ian Agol Aug 24 '11 at 16:32

1 Answer 1

up vote 6 down vote accepted

I believe it is still open. By the Boone-Higman Theorem (W. W. Boone and G. Higman, "An algebraic characterization of the solvability of the word problem", J. Austral. Math. Soc. 18, 41-53 (1974)), a finitely presented group has solvable word problem if and only if it can be embedded in a simple group that can be embedded in a finitely presented group.

It is widely believed that it is possible for the simple group itself to be finitely presented, but (AFAIK) not proved.

So the answer to Agol's comment is that no finitely presented group with unsolvable word problem can be embedded into a finitely presented simple group.

share|improve this answer
2  
Thanks, Derek! So, this question is really great then!! The thing about Boone-Higman Theorem is that it rests on Higman's Embedding Theorem, the proof of which is quite "non-constructive" (well, it is difficult to wait for some effective finite presentation when working with general recursively enumerable presentations), and when one tries to find a finite presentation for a simple group to embed a given presentation, there is needed some sort of "constructiveness". I guess there must be invented something completely new. –  Victor Aug 24 '11 at 17:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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