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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!

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    $\begingroup$ 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? $\endgroup$
    – Ian Agol
    Commented Aug 24, 2011 at 16:32

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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.

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    $\begingroup$ 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. $\endgroup$
    – Victor
    Commented Aug 24, 2011 at 17:58
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    $\begingroup$ @Victor: This is not true; the Higman Embedding Theorem is very constructive. Given a recursive presentation (finite set of generators, and Turing machine T which enumerates relators), you can explicitly construct a finitely presented group into which it embeds, as well as an explicit embedding. The number of generators and relators in this new group is linear in n and s (n = number of generators in original group, s = number of states of T) See J. Rotman, An introduction to the theory of groups, Springer-Verlag, New York, (1995), chapter 12. $\endgroup$
    – MCC
    Commented Mar 12, 2015 at 17:39
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There is a strengthening of the Boone-Higman result, due to Thompson. He showed that we can take the simple group to be finitely generated. In full, this reads:

"A finitely presented group has solvable word problem if and only if it can be embedded in a finitely generated simple group that can be embedded in a finitely presented group."

You can find the full details in: R. J. Thompson, "Embeddings into finitely generated simple groups which preserve the word problem", Word Problems II: The Oxford Book, Studies in Logic and the Foundations of Mathematics, Volume 95, (1980).

As far as I am aware, your original question "Does every finitely presented group with soluble word problem embed in a finitely presented simple group?" is still an open problem.

-Maurice

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