Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is the following:

Suppose that we are given the amalgamated product $ G = G_1 * _{G_3} G_2 $, and subgroups $ H_i \le G_i $ for $i=1,2,3$, such that in addition $H_3$ is as large as possible - namely, $H_3 = G_3 \cap H_1 \cap H_2 $.

Is is true that $ H = H_1 * _{H_3} H_2 $ embeds as a subgroup of $G$ in the obvious way, or it there some counter-example?

The standard results in this direction require, for example, that $H_1 \cap G_3 = H_2 \cap G_3$ , which of course might not hold in general.

share|cite|improve this question
I suspect you could make $H$ be an HNN-extension instead. Subgroups of free products with amalgamation were classified by Karrass and Solitar. See here. The paper is not an easy read, but it should help. – user1729 Aug 7 '13 at 19:40
@user1729 - thank you for the reference, I will definitely look into it. But - what do you mean, to make $H$ an HNN-extension? It is given as an amalgamated prodcut. (Did you mean that its image inside $G$ will become isomorphic to an HNN-extension)? – Guest Aug 7 '13 at 19:54
I have to say that I intuitively expect this to be true, I would be quite surprised to see a counter example. – Guest Aug 7 '13 at 19:56
Can't we just take $G_1 = \langle w,x,y \rangle$, $G_2 = \langle x,y,z \rangle$ both abelian (for example elementary abelian of order 8), $G_3 = \langle x,y \rangle$, $H_1 = \langle x \rangle$, $H_2 = \langle y \rangle$, $H_3 = 1$, then the subgroup generated by $H_1$ and $H_2$ is abelian, but $H_1 * H_2$ is not. – Derek Holt Aug 7 '13 at 21:37
@derek - Thanks - your nice example answers the question, and proves my intuition wrong :-) By the way, is there still a counter example if we let $G_1 = H_1$ ? – Guest Aug 8 '13 at 5:41

Your Answer


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

Browse other questions tagged or ask your own question.