Can this special amalgamated free product group be trivial group? Suppose $G$ is a nontrivial group, not necessarilly countable, and suppose we have two surjective group homomorphisms $\alpha, \beta: G\rightarrow G$, then we could form the ''amalgamated free product'' $H=G*_{G}G$ by using these two maps, see the precise definition of "amalgamated free product" here: http://en.wikipedia.org/wiki/Free_product#Generalization:_Free_product_with_amalgamation
then, is H always a nontrivial group? If not, then what else assumptions needed to make it is nontrivial.
 A: Yes it can be trivial. Consider groups and surjective homomorphisms $f:A\to B$, $g:A\to C$. Assume that the restrictions $f:Ker(g)\to B$ and $g:Ker(f)\to C$ are surjective too. Then the amalgam $D$ of $B$ and $C$ along $f$ and $g$ is trivial, because denoting the natural maps $u:B\to D$ and $v:C\to D$ we have $v(C)=v\circ g(Ker(f))=u\circ f(Ker(f))=1$ and similarly $u(B)=1$. Since $D$ is generated by $u(B)\cup v(C)$ is follows that $D$ is trivial. Now it's pretty clear how to find examples with $B,C$ both isomorphic to $A$ and nontrivial (hint: infinite direct sum, or so). 
A: I'm so sorry that my previous answer is completely wrong. I hope it didn't cause too much confusion.
Please check if the following argument makes sense.
By the universal property, for any group $W$, a morphism from $H=G*_GG$ to $W$ is just a pair of morphisms $(f,g)$ from $G$ to $W$ such that $f\circ \alpha=g\circ \beta$. Hence $H$ is trivial if and only if for all $W$ and for all pairs of morphisms $(f,g)$ from $G$ to $W$,
$f\circ \alpha=g\circ \beta$ implies that both $f$ and $g$ are trivial morphisms.
Now take an abelian group $A$ such that there is an epimorphism $i:A\to A\oplus A$. For example, we may take $A=\oplus_i^\infty \mathbb Z$. Now take $G=A\oplus A$ and $\alpha(a,b)=i(b)$, $\beta(a,b)=i(a)$. Then if $f\circ \alpha=g\circ \beta$ holds, they must both take $(a,b)$ to the identity. But this will imply that $f,g$ are trivial morphims since $i$ is surjective.
Best regards,
Yang
