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Higman's embedding theorem states every finitely generated, recursively presented group embeds into some finitely presented group. A further result of Higman, Neumann and Neumann shows that every countably generated, recursively presented group embeds into some finitely generated, recursively presented group.

I've read that a simple corollary of Higman's embedding theorem is that there exists a universal finitely presented group; one into which every finitely presented group embeds and therefore one into which every countably generated, recursively presented group embeds.

How does the existence of such a universal finitely presented group follow from the embedding theorem?

I thought that since there are countably many finitely presented groups (up to isomorphism), one could construct a group $G$ as the free product over an isomorphic copy of each finite presentation. This gives a countably generated group into which every finitely presented group embeds. So if the set of relators for $G$ is recursive then $G$ embeds into some finitely presented group and so that group would be universal. But the relator set for $G$ forms a countable union of finite sets and such unions aren't in general recursive or even recursively enumerable.

Is $G$ in fact recursively presented? If not, how does one prove the corollary?

Shaun
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h4tter
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    Can't you recursively enumerate all finite sesuences of relations ? – Maxime Ramzi May 05 '18 at 21:32
  • I'm not sure whether $G$ as stated is recursively presented. You may have to give up on having only one isomorphic copy of each finitely presented group as a factor. – Jalex Stark May 05 '18 at 21:47
  • There's no problem. You enumerate all finite sequences of words, you take the free product. You don't bother with which groups are gotten with these presentations: you'll get trivial groups plenty of times, nevermind. – YCor May 05 '18 at 22:43
  • "An isomorphic copy of each presentation" is imprecise (what's an isomorphism between presentations???). If one means one isomorphic copy of each finitely presented group, certainly there's a problem. So, don't bother: one copy of each presentation, period. – YCor May 05 '18 at 22:46
  • One can indeed consider an injective enumeration $(G_n)$ of all isomorphism classes of f.p. groups and consider the free product of all $G_n$ (or the restricted direct product). One can wonder whether the resulting group has a recursive presentation. My guess is no, but I don't know. – YCor May 05 '18 at 22:50
  • So if we take one copy of each presentation, we again get a countably generated group. But how do you enumerate all the finite sequences of words when they're over different generators? – h4tter May 05 '18 at 22:50
  • You enumerate all words in the infinite sequence $(x_n)$. (Alternatively, it's actually enough to consider words in 2 generators, relying on H-N-N.) – YCor May 05 '18 at 22:56
  • I'm still not sure I see why the resulting set of words is recursively enumerable. The way I'm thinking about the resulting language of relators is that it is the union of every finite language, but each of these finite languages is over a different alphabet. So are you saying we just pick an enumeration of the alphabets, then enumerate the words in each language, in the order of the enumeration of the alphabets? – h4tter May 05 '18 at 23:08
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    Oh, I think I see now. We can associate to each word a natural number. Then we can enumerate all the possible finite sequences of words by first enumerating all the sequences that contain only words with natural numbers less than 1, then enumerate all the sequences that contain only words with natural numbers less than 2, etc. So the free product of all finite presentations indeed has a recursively enumerable presentation. Then we can build a recursive presentation using Craig's trick. @YCor, if you'd like to write an answer I'd happily accept it. – h4tter May 05 '18 at 23:41

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