The Unapologetic Mathematician

Mathematics for the interested outsider

Amalgamated free products

I was thinking that one thing might have been unclear in my discussion of free products. The two groups in the product must be completely disjoint — there are no elements in common. Particularly, if we take the free product G*G we must use two different — but isomorphic — copies of G. For example, in S_3*S_3 we might take the first copy of S_3 to be permutations of \{1,2,3\}, and the second to be permutations of \{A, B, C\}.

So what if the groups aren’t disjoint? Say they share a subgroup. There’s a tool we can use to handle this and a few other situations: the amalgamated free product. Here’s the picture:

Universal Property of Pushouts

This diagram looks a lot like the one for free products, but it has the group H up at the top and homomorphisms from H into each of G_1 and G_2, and we insist that the square commutes. That is, we send an element of H into G_1 and on into G_1*_HG_2 or we can send it into G_2 and on into G_1*_HG_2, and the result will be the same either way. In many cases H will be a common subgroup of G_1 and G_2, and the homomorphisms from H are just the inclusions.

So how do we read this diagram? We start with three groups and two homomorphisms — G_1\leftarrow H\rightarrow G_2 — and we look for groups that “complete the square”. The amalgamated free product G_1*_HG_2 is the “universal” such group — given any other group X that completes the square there is a unique homomorphism from G_1*_HG_2 to X.

Just like for free products we’ve given a property, but we haven’t shown that such a thing actually exists. First of all there should be a homomorphism from G_1*G_2 to G_1*_HG_2 by the universal property of G_1*G_2. Then we have two ways of sending H into G_1*G_2: send H to G_1 sitting inside G_1*G_2 or send it to G_2 inside G_1*G_2. Let’s call the first way f_1:H\rightarrow G_1*G_2 and the second way f_2:H\rightarrow G_1*G_2. Now we want to add the relation f_1(h)=f_2(h) for each element of H. That is, f_1(h)f_2(h)^{-1} should be the identity. It isn’t the identity in G_1*G_2, but we can make it the identity by taking the smallest normal subgroup N of G_1*G_2 containing all these elements and moving to the quotient (G_1*G_2)/N. This is the group we’re looking for.

We’ve shown that (G_1*G_2)/N does complete the square. Now if X is any other group that completes the square it has homomorphisms into it from each of G_1 and G_2, so there’s a unique homomorphism f:G_1*G_2\rightarrow X. But now f sends every element of N to the identity in X because we assumed that X makes the outer square in the diagram commute. Since N is in the kernel of f we get a well-defined homomorphism from (G_1*G_2)/N to X, which is the one we need.

Amalgamated free products will become very important somewhere down the road, especially because they (and related concepts) figure very prominently in my own work.

March 3, 2007 Posted by | Algebra, Group theory, Structure of Groups, Universal Properties | 8 Comments

   

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