# Generalized free product of semigroups with amalgamated subsemigroups

Hanna Neumann in

[American Journal of Mathematics, 1948,

introduced a notion of generalized free product of groups $G_i$ with amalgamated subgroups $H_{ij}< G_i$. Did anybody consider this construction for semigroups? Thank you in advance.

This question arose while studying partial actions of groups on semigroups.

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Yes people have although most of the attention has been focused on one subsemigroup. –  Benjamin Steinberg Jul 27 '13 at 14:14
@ Benjamin Steinberg: Benjamin, thank you, but I am interested just in the case of several subsemigroups. –  Boris Novikov Jul 27 '13 at 14:28
@BorisNovikov: Boris, I know nothing about semigroups, but for groups, $\pi_1$ of a finite graph of groups can be always reduced to an iterated amalgam/HNN extension where on each step only one subgroup is involved. For countable graphs one can use this + direct limit. –  Misha Jul 27 '13 at 15:08
@Misha: Misha, spasibo. This information is useful for me also. –  Boris Novikov Jul 27 '13 at 15:42
@ Mark Sapir: Mark, all of these references deal with the case when there is only one amalgamated subsemigroup -- the intersection of all semigroups. I need the case when we have a family of amalgamated subsemigroups -- the intersections of pairs of semigroups. –  Boris Novikov Jul 27 '13 at 16:09

OK, here is my comment expanded to an answer. I looked at Hanna Neumann's definition again. What she defined is not a special case graph of groups, but of what is now called (again, in a special case) the "fundamental group of a complex of groups". Indeed, the modern definition of the latter is category-theoretic one (the limit of a commutative diagram of groups), details could be found for instance in the book by Bridson and Haefliger "Metric Spaces of Non-Positive Curvature". This definition extends verbatim to the case of semigroups. It is then well-known that vertex groups in such complex do not (in general) embed in the limit. However, they do provided that the complex is developable, which, in turn, is implied for instance by the local CAT(0) assumption.

Search for "complex of semigroups" returned nothing. My guess then is that you are on your own. My suggestion is to look for a CAT(0)-concept which might work in this context and would yield a "developable complex" of semigroups and guarantee, say, that vertex semigroups are embedded. Reading first Stallings' paper "Triangles of groups" (where "angles" are defined purely algebraically) might be a good start. See also here and Bridson-Haefliger.

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Thank you very much, Misha. –  Boris Novikov Aug 1 '13 at 16:51

The main difference between amalgamating groups and semigroups is that amalgamation is not always possible for semigroups. One of the simplest examples (from Lyapin's book) is the following.

Take two commutative semigroups (which are even groups): $$A=\{\dots,{1\over16},{1\over8},{1\over4},{1\over2},1,2,4,\dots\} \quad\hbox{and}\quad B=\{\dots,{1\over16},-{1\over8},{1\over4},-{1\over2},1,2,4,\dots\},$$ where the operation in $A$ is the usual multiplication and the operation in $B$ is $x\circ y=\pm xy$, where the sign is chosen so that $x\circ y\in B$.

It is easy to see that the amalgam (where $A\cap B=\{\dots,{1\over16},{1\over4},1,2,4,\dots\}$) cannot be embedded in a common semigroup. Indeed, in such a common semigroup, we would have $${1\over2}={1\over2}\cdot1={1\over2}\cdot(2\circ(-{1\over2}))= ({1\over2}\cdot2)\circ(-{1\over2})=1\circ(-{1\over2})=-{1\over2}.$$

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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. –  Andreas Blass Jul 27 '13 at 17:23
@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... –  Boris Novikov Jul 27 '13 at 17:44
@AndreasBlass, surely, you are right. –  Anton Klyachko Jul 27 '13 at 18:04
@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? –  Anton Klyachko Jul 27 '13 at 18:22
@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. –  Boris Novikov Jul 27 '13 at 19:35

If you allow many amalgamated subsemigroups, then the notion is too general. Every finitely presented semigroup can be presented that way with all $G_i$ free semigroups. For example, take any semigroup with one defining relation $S=\langle X\mid u=v\rangle$. Then consider a copy $X'$ of $X$, with a bijection $x\mapsto x'$, and two free semigroups $F(X), F(X')$. Identify subsemigroup $\langle x\rangle$ of $F(X)$ with $\langle x'\rangle$ of $F(X')$ for every $x\in X$, and $\langle u\rangle$ with $\langle v'\rangle$. Then you get $S$. Similarly, you can get any finitely presented semigroup, and if you allow infinite number of amalgamated subsemigroups - any semigroup.

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Thank you very much. But in general the semigroups $G_i$ don't embed into this construction, and first of all I am interested just by this topic. Of course, I understand that this topic is too complicated, so I didn't include it into my question. –  Boris Novikov Jul 28 '13 at 16:37
The problem whether $G_i$ embed into the product is undecidable. In fact the word problem is not necessarily decidable for amalgamated product of two finite semigroups with one amalgamated subsemigroup, even if you assume that both semigroups embed. It was proved by Kublanovsky and myself. –  Mark Sapir Jul 28 '13 at 16:48
I have a special case, so maybe... –  Boris Novikov Jul 28 '13 at 16:59
@Boris: The question was "where was it used". The answer is "nowhere" because the construction is too general. Neumann's construction is much more useful. But it is just a composition of amalgamated products with one amalgamated subgroup each. –  Mark Sapir Jul 28 '13 at 19:56
@ Mark Sapir: No, Mark. My question is "who consider". If nobody, we have to give definitions, establish simple properties etc. We have not "academic" interest, this construction appeared in our problems. Secondly, "it is just a composition of amalgamated products with one amalgamated subgroup each" - again, no. In the cited paper of Hanna Neumann several amalgamated subgroups were considered. –  Boris Novikov Jul 28 '13 at 20:52