Is the class of finite groups a Fraïssé class? Calling this class $K$, does $K$ satisfy the following:
- Joint embedding property
- Amalgamation property
- Hereditary property: if $G \in K$ and $H \le G$, then $H \in K$)
(1) holds because if $G, H \in K$, then $G \times H \in K$ as well. Obviously $G$ and $H$ are both subgroups of $G \times H$.
(2) seems to hold. Suppose $G_1, G_2 \in K$, with $H = G_1 \cap G_2$. Then the amalgamated free product $G_1 *_H G_2$ contains both $G_1$ and $G_2$ as subgroups. However, is it true that $G_1 *_H G_2$ is finite?
(3) holds because substructures of finite groups are again finite groups. By definition, substructures must contain $0$ and must be closed under the group operation. Furthermore, every element has an inverse, because they all have finite order (you will eventually reach the inverse by multiplying an element by itself successively). Associativity is universal, so also holds. So every substructure is also a (sub)group.
Is my reasoning correct? I would appreciate some kind of proof that amalgamated free products of finite groups are also finite, if this is true, or some counterexample if not, because I don't know much about amalgamated free products.
Just for information, an amalgam $A*_{C}B$ of finite groups with $C = A \cap B$ is finite if and only if $C = A$ or $C =B$ ( ie there is an inclusion between the groups $A$ and $B$)- this is proved in Serre's book "Trees".
On the other hand, if $A$ is a finite group of order $m$ and $B$ is a finite group of order $n,$ then (by Cayley's theorem) the symmetric group $S_{h}$ (with $h = m+n)$ is a finite group which contains isomorphic copies of $A$ and $B$ with a trivial intersection. The group generated by these isomorphic copies is an epimorphic image of the free product.