Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, example?)

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

share|improve this question
1  
Please try to add a suitable tag in addition to [homework]; in this case, [group-theory] seems pretty clear. –  Arturo Magidin Jun 21 '11 at 4:12
    
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 Lampret 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
1  
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
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.