Direct product of free groups 
Can a direct product of two free groups of rank $2$ be a non trivial amalgamated free product over $\Bbb Z$?

I read that it can't be done.
I tried to see what the edge stabilizers of the amalgamated tree might look like if such analysis was possible using the fact that if two hyperbolic elements commute (such as belonging to a different factor in the direct product), then their invariant axis would be the same but haven't made any progress.
Can you give me any hints on how to proceed?
 A: 
Let $F_1,F_2$ be non-cyclic finitely generated groups. Then $F_1\times F_2$ is not amalgamated product or HNN extension over a cyclic subgroup.

Let $F_1,F_2$ be non-cyclic finitely generated groups. Suppose that $G=F_1\times F_2$ acts unboundedly, inversion-free, on a tree $T$ with  cyclic edge stabilizers.
Suppose that each element of $F_1$ acts with a fixed vertex. Then $F_1$ has a fixed vertex (this is a general fact on tree actions of f.g. groups). Let $T_1$ be the set of vertices fixed by $F_1$. Then $T_1$ is a subtree, and is $F_2$-invariant. By minimality, we deduce that $T=T_1$, and hence $F_1$ acts trivially. Since $F_1$ is not cyclic, it follows that edge stabilizers are not cyclic, contradiction.
So some element $g_1$ of $F_1$ is loxodromic, with axis $A_1$. Similarly, some element $g_2$ of $F_2$ is loxodromic, with axis $A_2$. Since $F_2$ commutes with $g_1$, $F_2$ preserves $A_1$, and hence $A_2=A_1$, and since $F_1$ commutes with $g_2$, $F_1$ preserves $A_2=A_1$. So the action preserves an axis, hence by minimality $T$ is reduced to an axis. 
Hence $G$ modulo a cyclic normal subgroup is trivial, of order $2$, infinite cyclic, or dihedral. This is excluded by the assumptions.
