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A result of Higman states that there exists a finitely-presented group $G$ in which all other finitely-presented groups embed - I'll call such a group universal. Every countable group embeds in a 2-generated group, so there are 2-generated universal groups.

I was told that someone somewhere wrote down some explicit presentations of such 2-generated universal groups. Where can I find such presentations? Is the minimal number of relators for such a 2-generated universal group known? Lower bound?

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  • $\begingroup$ The OP is about two different classes of groups: 2-generated universal countable groups (these containas subgroups all countable groups and are not finitely presented) and universal finitely presented group (these contain all recursively presented groups as subgroups and are finitely presented). The question about small number of relators makes sense for the second class only. The only known lower bound is 1 (1-related groups have solvable word problem and are not universal). $\endgroup$
    – user6976
    Dec 2, 2019 at 23:05
  • $\begingroup$ The closest thing I know of in the literature is this paper of Chiodo and Hill: arxiv.org/abs/1610.00977 . They give an explicit procedure for embedding an arbitrary finitely presented group into a group with 8 generators and 26 relations. In principle, one could apply some HNN extensions to make this into a 2-generated group. As Mark has pointed out, this upper bound on the number of relations is a long way from the best available lower bound, namely 2. $\endgroup$
    – HJRW
    Dec 5, 2019 at 14:12

1 Answer 1

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As I wrote in my comment above, the OP is about two different classes of groups: 2-generated universal countable groups (these contain all countable groups and are not finitely presented) and universal finitely presented group (these contain all recursively presented groups and are finitely presented). The question about minimal number of relators makes sense for the second class only. The only known lower bound is 1 (1-related groups have solvable word problem and are not universal). Every universal finitely presented group has unsolvable word problem. I do not think there is an example of a universal group with few relators. The state of the art of few-relator groups with unsolvable word problem is in Collins paper of 1986: https://projecteuclid.org/download/pdf_1/euclid.ijm/1256044631. The smallest known presentation has 12 relators.

See also Panov, D. , Petrunin, A. The telescopic construction: a microsurvey. Proceedings of the Gökova Geometry-Topology Conference 2013, 110–119, Gökova Geometry/Topology Conference (GGT), Gökova, 2014.

for a different notion of universality of f.p. groups.

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  • $\begingroup$ Oh OK thanks for clearing that up - do you know of any explicit presentations for a f.p. group that contains all other f.p. groups? $\endgroup$
    – user101010
    Dec 3, 2019 at 10:54
  • $\begingroup$ There is no known explicit presentation. The easiest constructionn is this. Take the free group $F$ of rank 2. Consider an ininite set L of words satisfying small cancelation property $C'(1/12).$ For every finite set of relators$W$ written in an infinite alphabet consider the set of words $W_1$ obtained by replacing letters with words from L. Never use the same word from L for two different finite set of relators. Denote the union of the resulting finite sets of words by R. The group G=F/《R》 contain all countable recursively presented groups but is not f.p. ... $\endgroup$
    – user6976
    Dec 3, 2019 at 12:28
  • $\begingroup$ ...Take a Turing machine listing all telators from R. Use the construction from Rotman's group theort book to embed G into a f.p. group H. H is what you want. $\endgroup$
    – user6976
    Dec 3, 2019 at 12:29

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