Is there a universal countable group? (a countable group containing every countable group as a subgroup) 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?
 A: Hall's universal group is a countable locally finite group that contains every countable locally finite group (see these lecture notes).
A: 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.
A: 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.
A: It is, however, easy to see that there is a universal countable Abelian group. 
