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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?

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  • $\begingroup$ (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. $\endgroup$ – YCor Nov 23 '18 at 19:14
  • $\begingroup$ 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). $\endgroup$ – Geoffrey Janssens Nov 23 '18 at 22:07
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    $\begingroup$ 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. $\endgroup$ – YCor Nov 23 '18 at 22:12
  • $\begingroup$ Thanks for the clarification. Condition (a) and (b) is actually fully my setting, so I would be very interested in a positive answer. $\endgroup$ – Geoffrey Janssens Nov 23 '18 at 23:04
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    $\begingroup$ 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 $G$). $\endgroup$ – YCor Nov 30 '18 at 20:11
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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.

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

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