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Infinite finitely presented simple group (or more generally with trivial profinite completion) that is not amalgamated free product
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As is described in the title. Is there a finitely presented group $G$, with trivial profinite completion $widehatG=0$, which is not amalgamated free product?
For example, the famous example Higman groups are all amalgamated free product.
gr.group-theory geometric-group-theory
edited 18 mins ago
YCor
26.2k379124
asked 6 hours ago
Bruno
1307
add a comment |Â
up vote
3
down vote
favorite
As is described in the title. Is there a finitely presented group $G$, with trivial profinite completion $widehatG=0$, which is not amalgamated free product?
For example, the famous example Higman groups are all amalgamated free product.
gr.group-theory geometric-group-theory
edited 18 mins ago
YCor
26.2k379124
asked 6 hours ago
Bruno
1307
5
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
1
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
As is described in the title. Is there a finitely presented group $G$, with trivial profinite completion $widehatG=0$, which is not amalgamated free product?
For example, the famous example Higman groups are all amalgamated free product.
gr.group-theory geometric-group-theory
edited 18 mins ago
YCor
26.2k379124
asked 6 hours ago
Bruno
1307
As is described in the title. Is there a finitely presented group $G$, with trivial profinite completion $widehatG=0$, which is not amalgamated free product?
For example, the famous example Higman groups are all amalgamated free product.
gr.group-theory geometric-group-theory
gr.group-theory geometric-group-theory
edited 18 mins ago
YCor
26.2k379124
asked 6 hours ago
Bruno
1307
edited 18 mins ago
YCor
26.2k379124
asked 6 hours ago
Bruno
1307
edited 18 mins ago
YCor
26.2k379124
edited 18 mins ago
YCor
26.2k379124
edited 18 mins ago
YCor
26.2k379124
26.2k379124
asked 6 hours ago
Bruno
1307
asked 6 hours ago
Bruno
1307
asked 6 hours ago
Bruno
1307
1307
5
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
1
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago
add a comment |Â
5
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
1
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago
5
5
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
1
1
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
I'm going to flesh out my comment above to an answer. Thompson's groups $T$ and $V$ are famous examples of finitely presented infinite simple groups.
In this paper of Dan Farley, it is shown that $T$ and $V$ have Serre's property FA, which means that every time they act on a tree there is a global fixed point. (Farley says that this statement is originally due to Ken Brown.) When applied to the Bass--Serre tree of a splitting, it follows that any such splitting of $T$ or $V$ is trivial.
answered 5 hours ago
HJRW
17.9k250115
I see. Thanks for the explanation.
– Bruno
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
I'm going to flesh out my comment above to an answer. Thompson's groups $T$ and $V$ are famous examples of finitely presented infinite simple groups.
In this paper of Dan Farley, it is shown that $T$ and $V$ have Serre's property FA, which means that every time they act on a tree there is a global fixed point. (Farley says that this statement is originally due to Ken Brown.) When applied to the Bass--Serre tree of a splitting, it follows that any such splitting of $T$ or $V$ is trivial.
answered 5 hours ago
HJRW
17.9k250115
I see. Thanks for the explanation.
– Bruno
3 hours ago
add a comment |Â
up vote
7
down vote
accepted
I'm going to flesh out my comment above to an answer. Thompson's groups $T$ and $V$ are famous examples of finitely presented infinite simple groups.
In this paper of Dan Farley, it is shown that $T$ and $V$ have Serre's property FA, which means that every time they act on a tree there is a global fixed point. (Farley says that this statement is originally due to Ken Brown.) When applied to the Bass--Serre tree of a splitting, it follows that any such splitting of $T$ or $V$ is trivial.
answered 5 hours ago
HJRW
17.9k250115
I see. Thanks for the explanation.
– Bruno
3 hours ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
I'm going to flesh out my comment above to an answer. Thompson's groups $T$ and $V$ are famous examples of finitely presented infinite simple groups.
In this paper of Dan Farley, it is shown that $T$ and $V$ have Serre's property FA, which means that every time they act on a tree there is a global fixed point. (Farley says that this statement is originally due to Ken Brown.) When applied to the Bass--Serre tree of a splitting, it follows that any such splitting of $T$ or $V$ is trivial.
answered 5 hours ago
HJRW
17.9k250115
I'm going to flesh out my comment above to an answer. Thompson's groups $T$ and $V$ are famous examples of finitely presented infinite simple groups.
In this paper of Dan Farley, it is shown that $T$ and $V$ have Serre's property FA, which means that every time they act on a tree there is a global fixed point. (Farley says that this statement is originally due to Ken Brown.) When applied to the Bass--Serre tree of a splitting, it follows that any such splitting of $T$ or $V$ is trivial.
answered 5 hours ago
HJRW
17.9k250115
edited 5 hours ago
answered 5 hours ago
HJRW
17.9k250115
answered 5 hours ago
HJRW
17.9k250115
answered 5 hours ago
HJRW
17.9k250115
17.9k250115
I see. Thanks for the explanation.
– Bruno
3 hours ago
add a comment |Â
I see. Thanks for the explanation.
– Bruno
3 hours ago
I see. Thanks for the explanation.
– Bruno
3 hours ago
I see. Thanks for the explanation.
– Bruno
3 hours ago
add a comment |Â
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5
Thompson’s groups T and V are finitely presented infinite simple groups with Serre’s property FA; in particular, they don’t split.
– HJRW
6 hours ago
Any finite simple group clearly has property FA, and hence is not an amalgamated free product.
– Igor Belegradek
1 hour ago
1
@IgorBelegradek, that's true, but a finite simple group is equal to its profinite completion.
– HJRW
1 hour ago
Among infinite finitely presented simple groups, it is usually harder to to find non-FA groups than FA-groups. The first non-FA such groups are Burger-Mozes' groups. Checking that groups such as Thompson's group T or V have FA is quite easy. Also I'm not sure what is meant by "famous Higman groups". There's one famous Higman group (indeed an amalgam), generalizations (not called Higman groups as far as I know), and unrelated "Thompson-Higman groups" (which have FA).
– YCor
20 mins ago
It's always misleading when the question in the text does not match that in the title (of course the question can specify, but here it's the contrary, as simple groups are mentioned in the title and not the text).
– YCor
16 mins ago