Going up of an amalgamated decomposition of a subgroup of finite index
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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?
gr.group-theory graph-theory geometric-group-theory
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up vote
7
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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?
gr.group-theory graph-theory geometric-group-theory
(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 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 at 22:07
1
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 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 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
yesterday
add a comment |
up vote
7
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favorite
up vote
7
down vote
favorite
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?
gr.group-theory graph-theory geometric-group-theory
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?
gr.group-theory graph-theory geometric-group-theory
gr.group-theory graph-theory geometric-group-theory
asked Nov 23 at 18:03
Geoffrey Janssens
485
485
(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 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 at 22:07
1
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 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 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
yesterday
add a comment |
(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 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 at 22:07
1
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 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 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
yesterday
(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 at 19:14
(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 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 at 22:07
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 at 22:07
1
1
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 at 22:12
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 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 at 23:04
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 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
yesterday
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
yesterday
add a comment |
1 Answer
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7
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Yes, there are many examples: start from any group $A$, and consider $Awr C_2=A^2rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2subset Awr C_2.$$
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 $Awr C_2$ 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, $Awr F=A^Frtimes F$ 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 $Awr C_2$.
To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1rangle$), or many other examples.
The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $nge 2$.
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 $G=Awr F$ act on a tree $T$ without edge inversion. Write $A^F=prod_{iin F}A_i$.
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 $jneq i$. For $jneq i$, $D_j$ 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 $A_itomathrm{Aut}^+(D)simeqmathbf{Z}$. 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 at 22:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Yes, there are many examples: start from any group $A$, and consider $Awr C_2=A^2rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2subset Awr C_2.$$
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 $Awr C_2$ 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, $Awr F=A^Frtimes F$ 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 $Awr C_2$.
To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1rangle$), or many other examples.
The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $nge 2$.
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 $G=Awr F$ act on a tree $T$ without edge inversion. Write $A^F=prod_{iin F}A_i$.
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 $jneq i$. For $jneq i$, $D_j$ 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 $A_itomathrm{Aut}^+(D)simeqmathbf{Z}$. 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 at 22:08
add a comment |
up vote
7
down vote
accepted
Yes, there are many examples: start from any group $A$, and consider $Awr C_2=A^2rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2subset Awr C_2.$$
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 $Awr C_2$ 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, $Awr F=A^Frtimes F$ 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 $Awr C_2$.
To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1rangle$), or many other examples.
The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $nge 2$.
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 $G=Awr F$ act on a tree $T$ without edge inversion. Write $A^F=prod_{iin F}A_i$.
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 $jneq i$. For $jneq i$, $D_j$ 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 $A_itomathrm{Aut}^+(D)simeqmathbf{Z}$. 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 at 22:08
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Yes, there are many examples: start from any group $A$, and consider $Awr C_2=A^2rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2subset Awr C_2.$$
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 $Awr C_2$ 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, $Awr F=A^Frtimes F$ 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 $Awr C_2$.
To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1rangle$), or many other examples.
The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $nge 2$.
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 $G=Awr F$ act on a tree $T$ without edge inversion. Write $A^F=prod_{iin F}A_i$.
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 $jneq i$. For $jneq i$, $D_j$ 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 $A_itomathrm{Aut}^+(D)simeqmathbf{Z}$. But since $A$ has a finite abelianization, we deduce a contradiction.
Yes, there are many examples: start from any group $A$, and consider $Awr C_2=A^2rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2subset Awr C_2.$$
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 $Awr C_2$ 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, $Awr F=A^Frtimes F$ 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 $Awr C_2$.
To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1rangle$), or many other examples.
The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $nge 2$.
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 $G=Awr F$ act on a tree $T$ without edge inversion. Write $A^F=prod_{iin F}A_i$.
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 $jneq i$. For $jneq i$, $D_j$ 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 $A_itomathrm{Aut}^+(D)simeqmathbf{Z}$. But since $A$ has a finite abelianization, we deduce a contradiction.
edited Nov 23 at 19:11
answered Nov 23 at 18:39
YCor
27k380132
27k380132
Thanks a lot for your very clear and precise answer!! Also for the reference to your interesting article with Aditi.
– Geoffrey Janssens
Nov 23 at 22:08
add a comment |
Thanks a lot for your very clear and precise answer!! Also for the reference to your interesting article with Aditi.
– Geoffrey Janssens
Nov 23 at 22:08
Thanks a lot for your very clear and precise answer!! Also for the reference to your interesting article with Aditi.
– Geoffrey Janssens
Nov 23 at 22:08
Thanks a lot for your very clear and precise answer!! Also for the reference to your interesting article with Aditi.
– Geoffrey Janssens
Nov 23 at 22:08
add a comment |
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(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 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 at 22:07
1
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 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 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
yesterday