Amalgams of two nontrivial group is trivial I got this question in the book Tree by Serre. 
Let A=$Z$. $G_1=PSL(2,Q)$ and $G_2=Z/2Z$. We take $f_1: A\rightarrow G_1$ to be an 
injective (I do not know what is this injective map ?) and $f_2:\rightarrow G_2$ to be natural surjection map. Show that $G_1*_{A} G_2$=1$.
If I understand the injective map then I can play with the relation. Any help, hint will be very appreciated. 
 A: The amalgamation forces the copy of $A$ in $G_1$, $f_1(A)$, to be equal to the image of $A$ in $G_2$, $f_2(A)$. So we force $f_1(A)=f_2(A)$. As $f_1(A)$ is infinite cyclic but $f_2(A)$ is finite cyclic, this induces a homomorphism on $\operatorname{PSL}(2, \mathbb{Q})$. So it is a proper homomorphic image of $\operatorname{PSL}(2, \mathbb{Q})$ we get in $G_1\ast_AG_2$, it is not an embedding. For this example, the actual embedding $f_1: A\rightarrow G_1$ is irrelevant, because we just need the fact that we are inducing a proper homomorphic image.
As $\operatorname{PSL}(2, \mathbb{Q})$ is simple, the image of $G_1$ in $G_1\ast_AG_2$ (the induced homomorphic image) is trivial. So, writing $Tr$ for the trivial group, we have the following.
$$G_1\ast_AG_2\cong Tr\ast_AG_2=Tr\ast_AA=Tr$$
Which is what you are after.
Note that if the maps $f_1: A\rightarrow G_1$ and $f_2:A\rightarrow G_2$ are both injective then it is a theorem that the natural maps $G_1\rightarrow G_1\ast_AG_2$ and $G_2\rightarrow G_1\ast_AG_2$ are both injective. In fact, some authors stipulate that these maps must be injective$^{\dagger}$. For example, Lyndon and Schupp do, and I actually thought that Serre did!

$^{\dagger}$ Lyndon and Schupp go for something like: Given an an isomorphism $\phi$ between subgroups $A\leq G$ and $B\leq H$, the free product with amalgamation is the group with relative presentation $\langle G, H; a=\phi(a), a\in A\rangle$.
