How can I write $\mathbb{Z}^2 \ast \mathbb{Z}^2$ an HNN-extension with infinite cyclic associated subgroups? I have a free product  $\mathbb{Z}^2 \ast \mathbb{Z}^2$, and I need to express it as an HNN-extension with associated subgroups isomorphic to $\mathbb{Z}$.
I have not had much success. The way I am trying to think about it is that a free product of fundamental groups of spaces $X$ and $Y$ is the fundamental group of the wedge sum $X \vee Y$. So in this case, I have the wedge sum of two tori.
The HNN extension of a fundamental group of a space $Z$ is the gluing of a subgroup of the space to an isomorphic copy of that subgroup. Or something like that. So I've been trying to see how the wedge sum of two tori can be seen as the gluing of some space to itself along two isomorphic copies of $\mathbb{Z}$. This is kind of difficult to visualize and I'm not sure if I have the right idea.
Another related problem I have is expressing $\mathbb{Z}^2 \ast \mathbb{Z}^2$ as the fundamental group of a graph of groups, nontrivially, with 3 edges. I think here, nontrivially means I can't just add vertices or edges with trivial groups to pad the number of edges I have.
Obviously the trivial way is to have two vertices with group $\mathbb{Z}^2$ and an edge with trivial edge group connecting them. I cannot immediately see a nontrivial solution, since the only way I know to construct a graph of groups with a given fundamental group is to decompose it as some free products (with amalgamation) and HNN-extensions, then build up a graph using those.
If I had a solution to the first part, perhaps I could do some decomposition like:
$$\mathbb{Z}^2 \ast \mathbb{Z}^2 \cong X \ast_\mathbb{Z}$$
Which gives me 1 edge with group $\mathbb{Z}$, then I could decompose $X$ further in some way to get more edges. The problem is I don't know how to express $\mathbb{Z}^2 \ast \mathbb{Z}^2$ as an HNN-extension.
 A: You are over-thinking the HNN-part. $\mathbb{Z}^2$ splits as $$\mathbb{Z}^2\cong\langle a, t\mid t^{-1}at=a\rangle,$$ which is an HNN-extension with stable subgroups $\mathbb{Z}$. (This can be thought of topologically as taking a circle cross and interval and then glueing the end circles together.) This generalises to semidirect products $H\rtimes\mathbb{Z}$, and is called a mapping torus. Taking the free product doesn't affect this structure. We can actually do this in each of the factors to get:
$$\mathbb{Z}^2\ast\mathbb{Z}^2\cong\langle a, t_1, b, t_2\mid t_1^{-1}at_1=a, t_2^{-1}bt_2=b\rangle$$
This gives you a graph of groups with a single vertex and two loop edges. You can split this single vertex over the free product to get a graph of groups with two vertices and three edges, as required.
This counts as a "non-trivial" splitting: a "trivial" splitting is one where one of the vertex groups is the whole group ("is" rather than "is isomorphic to"). For example, let $A=\langle a\mid-\rangle$ and $B=\langle b\mid-\rangle$ (both are infinite cyclic). Then we can decompose $A$ as $A\ast_{a^2=b}B=\langle a, b\mid a^2=b\rangle$. This is a trivial splitting - it gives you no information about the group whatsoever!
