# If $\phi:G\twoheadrightarrow H$ and if $H=H_1*_{H_3}H_2$ for $H_i\leq H$, then is $G=G_1*_{G_3}G_2$, for $\phi(G_i)=H_i$?

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.

• Grushko-Neumann theorem/lemma could (I'm not sure) prove useful: Suppose $\varphi\!:F_r \rightarrow G\!=\!G_1\!\ast\!G_2$ is a surjective homomorphism, where $r\!=\!\mathrm{rank}(G)\!<\!\infty$. Then $F_r\!=\!F_{r_1}\!\ast\!F_{r_2}$ such that $\varphi(F_{r_i})\!=\!G_i$ for $i\!=\!1,2$. – Leon Jun 21 '11 at 5:24
• Surely the (other) trivial homomorphism will work, so $H=<1>$. Clearly $H=H_1\ast_{H_3}H_2$ where $H_i$ is trivial... – user1729 Jun 21 '11 at 9:59
• 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 $1$. – user641 Jun 22 '11 at 19:29