Mixing
A graph on which a free group acts freely is a tree?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?
If $F=mathbb Z$, the answer is no. For example $mathbb Z$ acts on $mathbb Z^2$ by translations.
But what about the other free groups?
As $F_2$ contains $F_k$ as a subgroup $forall k$, it would suffice to show that the claim is also falso for $F=F_2$.
However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1neq aba^-1$ acts as the identity.
abstract-algebra group-theory graph-theory
asked Jul 16 at 13:05
stacky
557
add a comment |Â
up vote
1
down vote
favorite
Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?
If $F=mathbb Z$, the answer is no. For example $mathbb Z$ acts on $mathbb Z^2$ by translations.
But what about the other free groups?
As $F_2$ contains $F_k$ as a subgroup $forall k$, it would suffice to show that the claim is also falso for $F=F_2$.
However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1neq aba^-1$ acts as the identity.
abstract-algebra group-theory graph-theory
asked Jul 16 at 13:05
stacky
557
3
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?
If $F=mathbb Z$, the answer is no. For example $mathbb Z$ acts on $mathbb Z^2$ by translations.
But what about the other free groups?
As $F_2$ contains $F_k$ as a subgroup $forall k$, it would suffice to show that the claim is also falso for $F=F_2$.
However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1neq aba^-1$ acts as the identity.
abstract-algebra group-theory graph-theory
asked Jul 16 at 13:05
stacky
557
Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?
If $F=mathbb Z$, the answer is no. For example $mathbb Z$ acts on $mathbb Z^2$ by translations.
But what about the other free groups?
As $F_2$ contains $F_k$ as a subgroup $forall k$, it would suffice to show that the claim is also falso for $F=F_2$.
However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1neq aba^-1$ acts as the identity.
abstract-algebra group-theory graph-theory
asked Jul 16 at 13:05
stacky
557
asked Jul 16 at 13:05
stacky
557
asked Jul 16 at 13:05
stacky
557
asked Jul 16 at 13:05
stacky
557
557
3
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37
add a comment |Â
3
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37
3
3
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853388%2fa-graph-on-which-a-free-group-acts-freely-is-a-tree%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Copying the example of $F = mathbb Z$, $F_2$ acts freely on the Cayley graph of $F_2 times H$ for any group $H$.
– Derek Holt
Jul 16 at 13:14
I think I got it. If I understood it correctly, it works for any grpah of the form $X times A$ wtih $A$ an arbitrary graph and $X$ the Cayley graph of $F_2$
– stacky
Jul 16 at 14:37