Understanding the free product amalgamation with an example I am trying to understand the $\ast -$product of two groups, I think I have managed to understand the free product of two arbitrary groups, but I am having problems when I consider $\ast_G$. The example I am trying to understand is $\mathbb{Z}^3 \ast_\mathbb{Z} \mathbb{Z}^3$, according to what I read I should consider $\varphi : \mathbb{Z} \to \mathbb{Z}^3$ and $\psi: \mathbb{Z} \to\mathbb{Z}^3$ (for example the inclusion?), and consider $\mathbb{Z}^3 \ast \mathbb{Z}^3$ but including the relations given by $\varphi(z)\psi^{-1}(z)=1$.
I don't understand how are this relations. Can you help me to understand this example?
 A: It's no harder to consider the general case. Suppose we are given the free product $F$ of the $\{G_i\}_{i\in I}$, an arbitrary group $A$, and monomorphisms $\{\alpha_i\}_{i\in I}$ from $A$ to the $G_i$.  We would like to construct another group, $G$, called the amalgamated free product, that satisfies a universal mapping property similar to the one that the free group does.
Look at the following diagram:

We want to complete the diagram, in such way that we can find our group $G$ and homomorphisms $f_i: G_i\to G$ such that the left square commutes, and such that if we are given homomorphisms $h_i:G_i\to H,$ another arbitrary group, as in the diagram, such that the outer square commutes, then there is a unique $h:G\to H$ making the two triangles commute.
Notice, whatever we do, we will need $f_i\circ\alpha_i(a)=f_j\circ\alpha_j(a)$, or what is the same thing, $f_i(\alpha_i(a))(f_j(\alpha_j(a))^{-1}=1$. The problem is this relation does not hold in the free product (there are no relations in the free product, by construction!). So we force the issue by taking $N$, the normal closure of $\text{all}$ these elements in the free product and then "modding out" $F$ by $N$: i.e. we take $G:=F/N.$
In your case, $I=\{1,2\},\ A=\mathbb Z,\ G_1=G_2=\mathbb Z^3.$ I'll leave it to you to say what the arrows are, and to check that the universal property holds.
