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If $G$ is a group, with epimorphism $\phi \colon G\rightarrow H$, and if $H=H_1*_{H_3}H_2$ for $H_i\leq H$, then is it true that $G=G_1*_{G_3}G_2$, where $\phi(G_i)=H_i$? If yes, how? If not, what would be a counterexample?

I tried to universal property of the amalgamated free product, but failed.

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    First note that the $G_i$ are uniquely determined (think of $H$ as $G/K$, where $K$ is the kernel of $\phi$), so the correspondence theorem shows that $G_1\cap G_2=G_3$. For what you say to be true, you need to show two things: (i) every element of $G-G_3$ can be written as an alternating product of elements from $G-G_1$ and $G-G_2$; and (ii) no such product is $1$.2011-06-22

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