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?
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Sign up to join this communityDoes there exist a group $G$ such that
If yes, is it the only one?
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.
The answer is yes if:
This group is known as the FraĂŻssĂ© 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