2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $i\!=\!1,2$. $\endgroup$ – Leon Jun 21 '11 at 5:24
  • $\begingroup$ Surely the (other) trivial homomorphism will work, so $H=<1>$. Clearly $H=H_1\ast_{H_3}H_2$ where $H_i$ is trivial... $\endgroup$ – user1729 Jun 21 '11 at 9:59
  • 2
    $\begingroup$ 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 $1$. $\endgroup$ – user641 Jun 22 '11 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy