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?



Going up of an amalgamated decomposition of a subgroup of finite index
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](https://arxiv.org/abs/1004.2582), [J. Group Theory publisher link](https://www.degruyter.com/view/j/jgth.2011.14.issue-1/jgt.2010.044/jgt.2010.044.xml)). 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 $n\ge 2$.




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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.


