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?