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Maximal order of a torsion element in a hyperbolic group
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Suppose that $G = langle X rangle$ is a $delta$-hyperbolic group, i.e. all geodesic triangles are $delta$-thin (the inverse image of a point under the projections onto a tripod has diameter bounded by $delta$).
It is a standard result that every finite subgroup of $G$ is conjugate to a subgroup contained in the ball of radius $2 delta +1$ around the identity. It follows that every finite subgroup of $G$ is of size at most $|X|^2delta + 1 + 1$.
Can this result be improved if we only considered cyclic subgroups, i.e. can we get a better bound on the maximal order of a torsion element than $|X|^2delta + 1 + 1$?
EDIT:
Following the discussion in the comments, does the situation change if we assumed the group $G$ to be one-ended?
group-theory hyperbolic-geometry
asked Jul 26 at 7:28
Michal Ferov
41229
add a comment |Â
up vote
2
down vote
favorite
Suppose that $G = langle X rangle$ is a $delta$-hyperbolic group, i.e. all geodesic triangles are $delta$-thin (the inverse image of a point under the projections onto a tripod has diameter bounded by $delta$).
It is a standard result that every finite subgroup of $G$ is conjugate to a subgroup contained in the ball of radius $2 delta +1$ around the identity. It follows that every finite subgroup of $G$ is of size at most $|X|^2delta + 1 + 1$.
Can this result be improved if we only considered cyclic subgroups, i.e. can we get a better bound on the maximal order of a torsion element than $|X|^2delta + 1 + 1$?
EDIT:
Following the discussion in the comments, does the situation change if we assumed the group $G$ to be one-ended?
group-theory hyperbolic-geometry
asked Jul 26 at 7:28
Michal Ferov
41229
This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose that $G = langle X rangle$ is a $delta$-hyperbolic group, i.e. all geodesic triangles are $delta$-thin (the inverse image of a point under the projections onto a tripod has diameter bounded by $delta$).
It is a standard result that every finite subgroup of $G$ is conjugate to a subgroup contained in the ball of radius $2 delta +1$ around the identity. It follows that every finite subgroup of $G$ is of size at most $|X|^2delta + 1 + 1$.
Can this result be improved if we only considered cyclic subgroups, i.e. can we get a better bound on the maximal order of a torsion element than $|X|^2delta + 1 + 1$?
EDIT:
Following the discussion in the comments, does the situation change if we assumed the group $G$ to be one-ended?
group-theory hyperbolic-geometry
asked Jul 26 at 7:28
Michal Ferov
41229
Suppose that $G = langle X rangle$ is a $delta$-hyperbolic group, i.e. all geodesic triangles are $delta$-thin (the inverse image of a point under the projections onto a tripod has diameter bounded by $delta$).
It is a standard result that every finite subgroup of $G$ is conjugate to a subgroup contained in the ball of radius $2 delta +1$ around the identity. It follows that every finite subgroup of $G$ is of size at most $|X|^2delta + 1 + 1$.
Can this result be improved if we only considered cyclic subgroups, i.e. can we get a better bound on the maximal order of a torsion element than $|X|^2delta + 1 + 1$?
EDIT:
Following the discussion in the comments, does the situation change if we assumed the group $G$ to be one-ended?
group-theory hyperbolic-geometry
asked Jul 26 at 7:28
Michal Ferov
41229
edited Jul 27 at 2:23
asked Jul 26 at 7:28
Michal Ferov
41229
asked Jul 26 at 7:28
Michal Ferov
41229
asked Jul 26 at 7:28
Michal Ferov
41229
41229
This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54
add a comment |Â
This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54
This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54
add a comment |Â
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This bound looks suspicious: finite cyclic groups are hyperbolic, with $delta(langle a | a^n rangle) = lfloor (n-1)/2 rfloor$, which is more than 1 + 1.
– xsnl
Jul 26 at 10:14
(nevermind, you consider symmetric generating sets)
– xsnl
Jul 26 at 10:15
In any case, if you consider $Bbb Z/2^k$ with presentation $langle x_1, dots, x_k-1 | x_i^2 = x_i+1, dots, x_k^2 = 1rangle$ or similar Cayley graph with low diameter, then exponent will be roughly exponential in triangle thinness.
– xsnl
Jul 26 at 10:43
@xsnl I was more thinking about infinite non-elementary case, but I think I see where you are going. Intuitively, a free product of a $delta_1$-hyperbolic and $delta_2$-hyperbolic group will be $maxdelta_1, delta_2$-hyperbolic, so taking a free product of a bunch of $mathbbZ/2^k$ (with the presentation you gave above), should be the kind of a infinite non-elementary hyperbolic group in which the maximal order of a torsion-element is exponential in with respect to the hyperbolicity constant.
– Michal Ferov
Jul 26 at 11:54