QUESTION: does there exist a group U such that three conditions hold:

  • (a) every finitely generated group is isomorphic to a subgroup of U;
  • (b) for every group G that is not finitely generated there exists a group H that is generated by G and one additional element x, and such that H is not isomorphic to any subgroup of U;
  • (c) (EDIT The following condition should not be here, please, forget it -- similar conditions would make sense for non-algebraic categories) U is generated by a countable set.


In the middle sixties (of the previous millennium) I posed a similar problem (almost a theorem :-) for compact metric spaces (instead of the finitely generated groups). In general, one can formulate similar problems for many different categories.

*Remark:  The question seems to make sense even in the case of finite group; this time even the third condition too is reasonable to consider.

  • 4
    $\begingroup$ — (c) means that $G$ is countable, but this obviously impacts the question whether or not you assume it. — As mentioned by J. Rickard, (a)+(c) is impossible. — (a) is obviously possible (the minimal cardinal of such a group has to be continuum). — (a)+(b) is not possible (regardless of (c)), just because every countable group embeds into a finitely generated group. $\endgroup$
    – YCor
    Sep 30 at 13:01

1 Answer 1



The third condition implies that $U$ is countable, and so has countably many finite subsets, and so has countably many finitely generated subgroups.

But there are uncountably many finitely generated groups, up to isomorphism, so the first condition contradicts the third.

  • $\begingroup$ right. I rushed from non-algebraic categories to algebraic. I will remove the third condition. $\endgroup$
    – Wlod AA
    Sep 30 at 12:36

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