This recent MO question, answered now several times over, inquired whether an infinite group can contain every finite group as a subgroup. The answer is yes by a variety of means.

So let us raise the stakes: Is there a countable group containing (a copy of) every countable group as a subgroup?

The countable random graph, after all, which inspired the original question, contains copies of all countable graphs, not merely all finite graphs. Is this possible with groups? What seems to be needed is a highly saturated countable group.

  • An easier requirement would insist that the group contains merely all finitely generated groups as subgroups, or merely all countable abelian groups. (Reducing to a countable family, however, trivializes the question via the direct sum.)

  • A harder requirement would find the subgroups in particularly nice ways: as direct summands or as normal subgroups.

  • Another strengthened requirement would insist on an amalgamation property: whenever $H_0\lt H_1$ are finitely generated, then every copy of $H_0$ in the universal group $G$ extends to a copy of $H_1$ in $G$. This property implies that $G$ is universal for all countable groups, by adding one generator at a time. This would generalize the saturation property of the random graph.

  • If there is a universal countable group, can one find a finitely generated such group, or a finitely presented such group? (This would lose amalgamation, of course.)

  • Moving higher, for which cardinals $\kappa$ is there a universal group of size $\kappa$? That is, when is there a group of size $\kappa$ containing as a subgroup a copy of every group of size $\kappa$?

  • Moving lower, what is the minimum size of a finite group containing all groups of finite size at most $n$ as subgroups? Clearly, $n!$ suffices. Can one do better?

  • $\begingroup$ I am sorry I downvote erroneausly since i am working with my phon. I can not not to correct it until the question would be edited. So I add a tag immediatly. I hope this tag is not far from your agreement. $\endgroup$ Commented Apr 17, 2018 at 7:08
  • $\begingroup$ @AliTaghavi No problem, please don't worry about it. $\endgroup$ Commented Apr 17, 2018 at 11:40

4 Answers 4


There isn't a countable group which contains a copy of every countable group as a subgroup. This follows from the fact that there are uncountably many 2-generator groups up to isomorphism.

The first example of such a family was discovered by B.H. Neumann. A clear account of his construction can be found in de la Harpe's book on geometric group theory.

  • $\begingroup$ Surely this is an immediate consequence of Higman, Neumann and Neumann's proof that you can embed any countable group in a 2-generator group? $\endgroup$
    – HJRW
    Commented Jun 21, 2010 at 21:52
  • $\begingroup$ Not quite ... an infinite 2-generator group often has uncountably many countable subgroups. So it seems necessary to prove that there are uncountably many finitely generated groups up to isomorphism. $\endgroup$ Commented Jun 21, 2010 at 21:58
  • $\begingroup$ Oh, right. Scratch that. I wasn't thinking straight. $\endgroup$
    – HJRW
    Commented Jun 21, 2010 at 22:03
  • 6
    $\begingroup$ Sorry for the late question about this answer, but why a f.g. group contains only countably many isomorphism classes of subgroups? $\endgroup$
    – Misha
    Commented Dec 15, 2013 at 21:02
  • 7
    $\begingroup$ @Misha sorry for the late response; a countable group has only countably many finite subsets, hence countably many finitely generated subgroups $\endgroup$ Commented May 15, 2019 at 16:30

No. There are uncountably many isomorphism classes of finitely generated groups, but a countable group contains only countably many finitely generated subgroups. There is a finitely presented group that contains all recursively presented groups, though.

I don't know how to prove that there are uncountably many isomorphism classes of finitely generated groups. I might try taking a finitely generated group that is not finitely presented and try the groups with the same generators and all subsets of the relations. If the example is too symmetric, like the lamplighter group, this probably won't work, but if you impose random relations of rapidly increasing length, it probably does work.

  • 1
    $\begingroup$ What I am curious about is whether there is an "easy" example of f.p. group containing just all finite groups. It seems that the "eventual shift permutations" group from the other thread isn't f.p. but perhaps the construction can be tweaked a bit to get one? And for the really ambitious, is there an $FP_\infty$ such group? $\endgroup$ Commented Jun 22, 2010 at 5:51
  • 2
    $\begingroup$ @Victor: The Thompson-Higman group $V$ is an example of a finitely presented group containing all finite groups. $\endgroup$
    – user6976
    Commented Sep 14, 2010 at 23:37

It is, however, easy to see that there is a universal countable Abelian group.

  • $\begingroup$ Peter, I guess you mean a universal countable abelian group. But could you explain a bit more? $\endgroup$ Commented Jun 22, 2010 at 13:08
  • 9
    $\begingroup$ Any countable abelian group can be embedded in a divisible countable abelian group. So you obtain a universal countable abelian groups by taking a direct sum of countably many copies of $\mathbb{Q}$ and $\mathbb{Z}[p^{\infty}]$ for each prime $p$. $\endgroup$ Commented Jun 22, 2010 at 13:35
  • $\begingroup$ Yes, Joel, that's what I meant. Thank you. $\endgroup$ Commented Jun 22, 2010 at 16:01

Hall's universal group is a countable locally finite group that contains every countable locally finite group (see these lecture notes).

  • $\begingroup$ This is really an answer to the earlier question, mathoverflow.net/questions/28945/… $\endgroup$ Commented Jun 22, 2010 at 11:31
  • $\begingroup$ Nope. Not every countable locally finite group containing all finite groups contains all countable locally finite groups. $\endgroup$
    – Someone
    Commented Jun 22, 2010 at 11:44
  • 1
    $\begingroup$ Wikipedia link only mentions that it contains all $\textit{finite}$ groups. With the stronger conclusion, I agree with you. $\endgroup$ Commented Jun 22, 2010 at 23:03

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