Timeline for Generalized free product of semigroups with amalgamated subsemigroups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2013 at 12:22 | comment | added | Boris Novikov | @Anton Klyachko: Certainly, I am agree with you. | |
| Jul 28, 2013 at 11:18 | comment | added | Anton Klyachko | @BorisNovikov, This is all right but my point is that it is much easier to answer more exact questions, something like "Is it true that...? Note that a similar statement for groups is true [HN48]." | |
| Jul 27, 2013 at 19:35 | comment | added | Boris Novikov | @Anton Klyachko: My question is simple. In our work we met the product of semigroups with amalgamated subsemigroups. We need results for it similar to results of H.Neumann for groups. We don't want "to reinvent the wheel" (изобретать велосипед), so I wonder whether anybody study such a product. | |
| Jul 27, 2013 at 18:22 | comment | added | Anton Klyachko | @BorisNovikov, yes, I do not fully understand the question. Note that even for groups, if we have 3 groups $G_i$ with 3 subgroups $H_{ij}=G_i\cap G_j$, this amalgam may be non-embeddable in any common group. There is, however, the notion (due to Bass and Serr) of the fundamental group of a graph of groups, but this is not an amalgamated product, this is a composition of amalgamated products and HNN-extensions (see, Misha's comment). What is your question, actually? | |
| Jul 27, 2013 at 18:04 | comment | added | Anton Klyachko | @AndreasBlass, surely, you are right. | |
| Jul 27, 2013 at 17:44 | comment | added | Boris Novikov | @Anton Klyachko: The problem of embedding an amalgame with only one amalgamated subsemigroup is (more or less) solved in Clifford-Preston, chapt.9. But my question is another... | |
| Jul 27, 2013 at 17:23 | comment | added | Andreas Blass | A category-minded person might define "amalgamation of $A$ and $B$ over $C$" as the universal example of a $G$ with homomorphisms from $A$ and $B$ that coincide on $C$. In this sense, the amalgamation is always possible (by the adjoint functor theorem or by an explicit presentation), but, as your example shows, the images of $A$ and $B$ in $G$ may intersect in more than just the image of $C$, and so the non-category-minded would be reluctant to accept this as an amalgamation. | |
| Jul 27, 2013 at 17:01 | history | answered | Anton Klyachko | CC BY-SA 3.0 |