Is there a smallest group containing all finite groups? Does there exist a group $G$ such that

*

*for any finite $K$ there is a monomorphism $K \to G$

*for any $H$ with property 1 there is a monomorphism $G \to H$
If yes, is it the only one?
 A: No. To show that it doesn't exist it is enough to produce two groups $G,H$ which contain isomorphic copies of all finite groups, but such that no group containing isomorphic copies of all finite groups embeds into both $G$ and $H$.
Let $(G_n)$ be an enumeration of all finite groups. Let $G=\bigoplus G_n$ be the restricted direct sum and  $H={\Large\ast}_nG_n$ the free product.
If $K$ is a subgroup of $G$ then $K$ is locally finite, hence freely indecomposable. Hence if $K$ is also isomorphic to a subgroup of $H$, then by Kurosh's subgroup theorem, $K$ is finite. In particular, $K$ doesn't contain isomorphic copies of all finite groups.
A: The answer is yes if:

*

*you furthermore impose that every finitely generated subgroup of $G$ is finite,
and

*you replace your requirement that the group be a smallest group containing all finite groups by the requirement that the group be a largest group containing all finite groups (within the class of group with the property that every finitely generated subgroup is finite).

This group is known as the Fraïssé limit of the category of finite groups.
See:
https://en.wikipedia.org/wiki/Fra%C3%AFss%C3%A9_limit 
and
https://math.stackexchange.com/questions/88169/fra%C3%AFss%C3%A9-limits-and-groups
