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Is the class of finite groups a Fraïssé class? Calling this class $K$, does $K$ satisfy the following:

  1. Joint embedding property
  2. Amalgamation property
  3. 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.


marked as duplicate by Martin, Amzoti, Julian Kuelshammer, Lord_Farin, Jim Jun 14 '13 at 7:33

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  • $\begingroup$ i don't think (2) should hold. a simple example (if it's allowed) is to take $G$ and $H$ as distinct copies of $\mathbb{Z}/2\mathbb{Z}$. then the intersection is trivial, and it's easy to see that the free product is infinite $\endgroup$ – citedcorpse Jun 14 '13 at 5:43
  • $\begingroup$ Oh yeah duh, that's the infinite dihedral group. In that case, is there a way to construct a finite group that contains both $G_1$ and $G_2$? The amalgamated free product fails, but I'm still not convinced that (2) does not hold. $\endgroup$ – Vikram Saraph Jun 14 '13 at 5:46
  • $\begingroup$ for example, it seems like $G_1$ and $G_2$ should appear in a large enough symmetric group, though I can't immediately see that this is the case. $\endgroup$ – Vikram Saraph Jun 14 '13 at 5:56
  • $\begingroup$ Never mind, I found my answer here: math.stackexchange.com/questions/88169/… $\endgroup$ – Vikram Saraph Jun 14 '13 at 5:59
  • $\begingroup$ ah excellent, it seems your intuition was right. i was trying to decipher the wikipedia page and found it confusing (i've never seen that language before). good work :) $\endgroup$ – citedcorpse Jun 14 '13 at 6:02

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.


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