# Amalgamated Free Product

Suppose $H$ is embedded in $G$ and $H'$ is isomorphic to $H$ and embedded in $G'$. Then we can simultaneously embed $H$, $H'$, $G$ and $G'$ into a single object (the amalgamated free product) such that $H$ and $H'$ become identifiable. This seems similar to the Isomorphism Extension Theorem for fields which is important for developing Galois Theory. What are the practical uses of the amalgamated free product?

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van Kampen's theorem says $\pi_1(A\cup B)$ is the free product of $\pi_1(A)$ and $\pi_1(B)$, amalgamated along $\pi_1(A\cap B)$. – user641 Mar 2 '11 at 19:43
A theorem of G. Baumslag says that the free product of two free groups, amalgamated along a cyclic subgroup, is residually finite. This for instance implies all surface groups are residually finite. – user641 Mar 2 '11 at 19:54
@Steve: With conditions on $A$, $B$, and $A \cap B$, of course. More generally it allows us to consider concretely what a "the most general group containing two subgroups intersecting in a particular way" looks like. The notion has a further extension in terms of a pushout, where the "intersection" need not be isomorphic in the two summands. – Zhen Lin Mar 2 '11 at 19:56
A theorem of J.R. Stallings says that a f.g. group with infinitely many ends is either (a special type of) an HNN extension or an amalgamated free product. – user641 Mar 2 '11 at 19:56

## 2 Answers

As Steve points out, amalgamated free product are key to the Seifert van-Kampen's theorem.

Other uses:

• HNN Extensions use amalgamated free products. HNN extensions are as follows: given a group $G$ and two subgroups $H$ and $K$ that are isomorphic as groups but not conjugate in $G$, one constructs and overgroup $\mathcal{G}$ of $G$ in which $H$ and $K$ are conjugate. Some consequences of HNN extensions: every countable group can be embedded in a 2-generated group; there exist infinite simple torsion-free groups; there are uncountably many isomorphism classes of finite generated groups; etc.

• Embeddability of amalgams. Suppose that in your setting $G$ and $G'$ satisfy some further group-theoretic property (e.g., they are finite, torsion, nilpotent, $p$-groups, solvable, etc). Can you find a group $\mathcal{G}$ that contains both $G$ and $G'$ as subgroups, in which $G\cap G' = H=H'$ (or at least, $G\cap G'\supseteq H=H'$), and $\mathcal{G}$ satisfies the same group theoretic property as $G$ and $G'$? Here the answer depends on the group theoretic property; for example, if $G$ and $G'$ are $p$-groups, you may be unable to find a $p$-group into which you can embed them with the properties above. Finding conditions for embeddability is important in terms of separation properties of morphisms, and the concept of epimorphism in special categories of groups (especially varieties).

• Completing "partial groups". Often one has a structure in which only some products are defined, and one wants to extend it to a group in which all products are defined so that you get a group. The amalgamated free product is the key to this.

• They are the group-theoretic categorical pushout of an inclusion diagram. Important in dealing with some notions of category theory within groups.

Added. Let me add that I do not think your analogy to the Isomorphism Extension Theorem is very apt, at least not if by "Isomorphism Extension Theorem" you are thinking about the same theorem as I am. By the IET I understand the following theorem:

Theorem. If $F$ is a field, $E$ is an algebraic extension of $F$, and $f\colon F\rightarrow K$ is a field embedding, then $f$ extends to an embedding $\overline{f}\colon E\to \overline{K}$ of $E$ into the algebraic closure of $K$.

I think you are thinking about the identification between $H$ and $H'$ as playing the role of the map $f$ above; but in the IET you already have a fixed "overfield" into which $K$ is embedded in a particular way, and into which you map $E$. Moreover, the extension $\overline{f}\colon E\to \overline{K}$ may "overlap" with $K$ in more than $f(F)$; that is, $\overline{f}(E)\cap K$ may be strictly larger than $f(F)$.

It would be more analogous to this if what you had was a group $G$ with a subgroup $H$, an embedding of $H$ into $G'$, and a group $\overline{G'}$, independent of $G$, into which $G$ could then be embedded extending the the map $H\hookrightarrow G'$.

I think that if you want to find an analogous theorem in fields, it would be that given a field $F$ and two algebraic extensions $K$ and $L$, there exists an extension $M$ of $F$ into which you can embed $K$ and $L$ (in such a way that $K\cap L\supseteq F$), and which is "universal" in the sense that any field which contains an image of $L$ and of $K$ will contain an image of $M$ (this field, of course, is the compositum of $L$ and $K$ in the algebraic closure of $F$).

This makes a bit more sense in that both are instances of pushouts, even though in the field case you need not have a strong embedding (one in which the intersection of the two objects is exactly their "common" object, and not something larger).

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Another important use of amalgamated products arises from groups acting on trees. One easy way to interpret $G=\mbox{SL}_2(\mathbb{Z})$ as an amalgam is to observe how it naturally acts on a tree embedded in the upper half plane $H$ (the tree being simply the boundary of the tessellation of $H$ by fundamental domains of the action of $G$).

See Serre's "Trees" ("Arbres, Amalgames, SL_2"), or these nice lecture notes.

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