# Going up of an amalgamated decomposition of a subgroup of finite index

Let $$G$$ be a finitely presented group and H a subgroup of index $$n$$ in $$G$$. Suppose that H has a non-trivial decomposition as amalgamated product, say $$H = A \ast_U B$$. I am wondering about the following two questions:

(i) If $$n = 2$$, does then $$G$$ also have a non-trivial amalgamated decomposition?

(ii) More generally, does there exists sufficient conditions (for example on $$A,B,U$$) such that $$G$$ also have non-trivial amalgamated decomposition?

• (ii) is a bit open-ended but at least one important case is when $H$ has infinitely many ends and has finite abelianization (maybe unnecessary): then $G$ also has infinitely many ends and one can use Stallings' theorem. – YCor Nov 23 '18 at 19:14
• Thanks! Your comment was already very enlightening to me (I was not truly familiar with theory of ends). Following some reference, e(H)= e(G) in general (so also without finite abelianization). – Geoffrey Janssens Nov 23 '18 at 22:07
• Yes but it's not the point. If $e(H)=\infty$, we deduce that $e(G)=\infty$, so by Stallings, $G$ decomposes as HNN or amalgam over a finite group. If $\mathbf{Z}$ is not a quotient of $H$, then it's not a quotient of $G$ and hence we can exclude the HNN possibility and hence $G$ splits as amalgam over a finite group. But just knowing that (a) $G$ has infinitely many ends (b) $G$ has some finite index subgroup with a nontrivial amalgam decomposition, it's unclear to me if $G$ also has a nontrivial amalgam decomposition. – YCor Nov 23 '18 at 22:12
• Thanks for the clarification. Condition (a) and (b) is actually fully my setting, so I would be very interested in a positive answer. – Geoffrey Janssens Nov 23 '18 at 23:04
• Possibly this could be asked separately: let $G$ be a group with an amalgam decomposition over a finite subgroup: does every finite index over group also have such an amalgam decomposition? I guess the answer is yes (without extra assumption on $G$). – YCor Nov 30 '18 at 20:11

Yes, there are many examples: start from any group $$A$$, and consider $$A\wr C_2=A^2\rtimes C_2$$, $$C_2$$ permuting both copies. Consider the inclusion of index 2 $$A^2\subset A\wr C_2.$$

a) If $$A$$ has a nontrivial amalgam decomposition, so does the group $$A^2$$ (since it has $$A$$ as quotient group).

b) It remains the question when $$A\wr C_2$$ has Property FA. This is answered in Theorem 1.1 of my paper with Aditi Kar (arxiv link, J. Group Theory publisher link). Assuming that $$A$$ is countable, and $$F$$ a nontrivial finite group, $$A\wr F=A^F\rtimes F$$ has Serre's Property FA if and only if $$A$$ is finitely generated and has finite abelianization. ($$\star$$)

So, for (a) let's assume that $$A$$ has a nontrivial amalgam decomposition, and for (b) let's assume that $$A$$ is finitely generated with finite abelianization. In this case, $$A^2$$ admits a nontrivial amalgam decomposition, but not its overgroup of index two $$A\wr C_2$$.

To satisfy both fulfillments, one can take $$A$$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $$\langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1\rangle$$), or many other examples.

The same holds with the cyclic group $$C_2$$ replaced with $$C_n$$ for arbitrary $$n\ge 2$$.

To be self-contained here's a proof of ($$\star$$). Let $$A$$ be a finitely generated group with finite abelianization, $$F$$ a nontrivial finite group, and let $$G=A\wr F$$ act on a tree $$T$$ without edge inversion. Write $$A^F=\prod_{i\in F}A_i$$.

Suppose that the action is unbounded. Then it is unbounded on $$A_i$$ for some $$i$$, and hence on each $$A_i$$ since they are all conjugate. Since $$A$$ is finitely generated, choose some element $$f_i$$ in $$A_i$$ acting as a loxodromic isometry; its axis $$D_i$$ is then invariant by $$A_j$$ for all $$j\neq i$$. For $$j\neq i$$, $$D_j$$ is $$f_j$$-invariant and hence $$D_j=D_i$$. This proves (using that $$|F|\ge 2$$ to ensure the existence of $$j$$) that $$D_i$$ is $$A_i$$-invariant. Also it does not depend on $$i$$; call it $$D$$: $$D$$ is the unique minimal $$A_i$$-invariant subtree $$T_i$$ of $$T$$. Since $$F$$ permutes the $$T_i$$, it thus preserves the axis $$D$$.

Since the actions of each $$A_i$$ on the axis $$D$$ commute with each other and are all unbounded and $$|F|\ge 2$$, they cannot be orientation reversing (since the centralizer of an orientation-reserving unbounded action on the line is trivial); so the action of $$A_i$$ on $$D$$ is given by a nontrivial homomorphism $$A_i\to\mathrm{Aut}^+(D)\simeq\mathbf{Z}$$. But since $$A$$ has a finite abelianization, we deduce a contradiction.

• Thanks a lot for your very clear and precise answer!! Also for the reference to your interesting article with Aditi. – Geoffrey Janssens Nov 23 '18 at 22:08