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.

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It is a standard result that for any free product $K = G*_AH$ of groups $G,H$ with amalgamated subgroup $A$, the natural maps $G \to K$ and $H \to K$ are injective. So $K$ can only be trivial if both $G$ and $H$ are trivial. – Derek Holt Oct 19 '12 at 9:02
@Holt, could you give an explaination why the natural map $G\rightarrow K$ is injective? since in our example, $H=G$, so if you consider the element $\alpha(g)\beta(g)^{-1}\in G$, it is mapped to identity in $G*_{G}G$ under the natural map, but $\alpha(g)\beta(g)^{-1}\in G$ is not necessarily the identity in $G$. – ougao Oct 19 '12 at 12:26
Sorry, I have to qualify my previous comment. I am more familiar with a definition of free product with amalgamation that involves an isomorphism between subgroups of the two groups. So what I said would be correct if $\alpha$ and $\beta$ were both isomorphisms. But I don't know what happens otherwise (which could only occur for a nonHopfian group $G$). – Derek Holt Oct 19 '12 at 13:34
@Derek, my first example is not correct to disprove the natural map $G\rightarrow K$ is injective, since the $\beta(g)^{-1}$ is in the second copy of $G$, so $\alpha(g)\beta(g)^{-1}\in G*G$; anyway, since according to Yves's answer below, the natural map could be trivial. – ougao Oct 19 '12 at 23:36

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).

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thanks for this nice idea. By the way, I asked this problem because I tried to use Van-kampen's theorem to prove the fundamental group the double of the cone on the Hawaiian Earring (gluing of two mapping cone of the Hawaiian earring along the "bad point") is nontrivial. According to your answer, maybe I need to seek other ways. – ougao Oct 19 '12 at 23:39
But the amalgamated free product in Van Kampen's theorem is 'injective', right? So then this problem can't occur. – Tara B Oct 20 '12 at 11:20
@Tara, this is not correct, since $U\cap V\subset U$ does not imply $\pi_1(U\cap V)\rightarrow \pi_1(U)$ is injective. You can check the definition of "Semi-locally simply connected spaces" to see the special case when the image is trivial. – ougao Oct 20 '12 at 17:25
@ougao: Sorry, yes, you're right of course. I hadn't thought about it properly. – Tara B Oct 20 '12 at 23:51

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

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But $G*_G G$ is defined as a quotient of $G*G$, not of $G$. – Derek Holt Oct 19 '12 at 9:04
I don't care about the explicit construction. The universal property characterizes this object uniquely. The explicit construction only proves existence. – Yang Zhou Oct 19 '12 at 15:06
I am so sorry that I am wrong. This is not the universal property of coequalizer. – Yang Zhou Oct 19 '12 at 15:18
@DerekHolt: I have modified my answer. Please help me check whether it makes sense this time. Sorry for my carelessness again! – Yang Zhou Oct 19 '12 at 15:47
Yes, it's OK now! – Derek Holt Oct 19 '12 at 21:31