# The Unapologetic Mathematician

## 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:

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