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I was recently reading up on Fraïssé limits in Hodges' "A shorter model theory." I was trying to think of some examples and wanted to see if I could take the Fraïssé limit on the category of finite groups. It is clear that this category has the hereditary and joint embedding property. However, I am having trouble showing that this category has the amalgamation property, by which I mean that if there are finite groups $G, H, K$ such that there are embeddings $\varphi: G\hookrightarrow H$ and $\psi: G\hookrightarrow K$, then there exists a finite group $J$ and embeddings $\vartheta: H\hookrightarrow J$ and $\eta: K\hookrightarrow J$ such that $\vartheta\circ \varphi = \eta\circ \psi$.

One simplification is that we can, by taking isomorphic copies if necessary, assume that $H\cap K = G$, and that $\varphi $ and $\psi$ are just inclusion maps. I am not even sure what group $J$ should be. Any help would be appreciated.

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It is a theorem of Schreier that every amalgam of two finite groups is strongly embeddable in a finite group. I think the result is contained in Die Untergruppen der freien Gruppen (Abh. math. Sem. Hamburg Univ. 5 (1927), 161-183); that paper at any rate contains the fact that any amalgam of two groups is strongly embeddable.

You could conclude it as an easy corollary of the following more recent theorem:

Theorem Let $B$ be a subgroup of both $K$ and $H$, and let $G$ be a subgroup of $B$ that is normal in $B$, $K$, and $H$, and such that $K/G$ and $H/G$ are finite. Then there is a strong amalgam of $K$ and $H$ over $B$ with $G$ normal and $J/G$ finite if and only if $K/GC_{K}(G)$ and $H/GC_H(G)$ generate a finite subgroup of the outer automorphism group of $G$.

Then apply the theorem with $G$ the trivial group.

This theorem appears in An amalgamation theorem for finite group extensions, by Frieder Haug, Ulrich Meierfrankenfeld, and Richard E. Phillips, Arc. Math. 57 (1991), p 325-331.

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  • $\begingroup$ Wish I could +1 again. I just found my notes on completing amalgams, and realized I used to call these Schreier embeddings. I came to update my answer, but found you already covered it. $\endgroup$ – Jack Schmidt Dec 7 '11 at 18:32
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There is also a fairly explicit construction of one such J: one takes G = HK, and transversals S and T of G in H and K, and then J is the subgroup of Sym( G × S × T ) generated by the regular representations of both H and K (where an element of H fixes each point of T and views G × S as H).

This typically creates very large J.

Any such J as you asked for is called a (faithful, finite) completion of the (rank 2) amalgam (H, K).

Choosing different transversals can create different (faithful, finite) completions J. However, I believe there are many more (faithful, finite) completions than those created by this method. All of them are finite quotients of the universal completion mentioned in the comments, the free product with amalgamation. Some have used the low index subgroup iterator to find finite faithful completions that have low index core-free subgroups.

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  • $\begingroup$ Thank you, this is very helpful. $\endgroup$ – student555 Dec 8 '11 at 8:50

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