Show $C subset G$ is relatively convex iff $G/C$ is left-orderable.
Definition: Given a left-orderable group $G$ with a left ordering $<$, a subgroup $H$ of $G$ is convex with respect to $<$ if for any elements $a<b in H$ and $a<c<b$ then $c in H$. A subgroup $Hsubset G$ is called relatively convex if there is an ordering in which $H$ is convex with respect to that ordering.
Prove: Suppose that $C$ is a subgroup of $G$, denote the set of left cosets ${gC}_{gin G}$ by $G/C$. The subgroup $C$ is relatively convex in $G$ if and only if there exists an ordering $prec$ of the cosets $G/C$ that is invariant under left multiplication by $G$, i.e. $gC prec hC$ implies $fgC prec fhC$ for all $f,g,h in G$.
Here is the question:
Here is problem 1.8 mentioned in the question above:
Here is my attempt:
Suppose $C$ is relatively convex, then there is an ordering $<$ on $G$ with respect to which, $C$ is convex. Then we define the ordering on the cosets of $C$ by $gC<hC$ if $g<h$. Now we need to show $<$ is left invariant and well defined. We first show that it is well defined: so suppose $g=g'c_1$, $h=h'c_2$, $g<h$ and $c_1<c_2$.(complete later)
abstract-algebra group-theory order-theory
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
add a comment |
Definition: Given a left-orderable group $G$ with a left ordering $<$, a subgroup $H$ of $G$ is convex with respect to $<$ if for any elements $a<b in H$ and $a<c<b$ then $c in H$. A subgroup $Hsubset G$ is called relatively convex if there is an ordering in which $H$ is convex with respect to that ordering.
Prove: Suppose that $C$ is a subgroup of $G$, denote the set of left cosets ${gC}_{gin G}$ by $G/C$. The subgroup $C$ is relatively convex in $G$ if and only if there exists an ordering $prec$ of the cosets $G/C$ that is invariant under left multiplication by $G$, i.e. $gC prec hC$ implies $fgC prec fhC$ for all $f,g,h in G$.
Here is the question:
Here is problem 1.8 mentioned in the question above:
Here is my attempt:
Suppose $C$ is relatively convex, then there is an ordering $<$ on $G$ with respect to which, $C$ is convex. Then we define the ordering on the cosets of $C$ by $gC<hC$ if $g<h$. Now we need to show $<$ is left invariant and well defined. We first show that it is well defined: so suppose $g=g'c_1$, $h=h'c_2$, $g<h$ and $c_1<c_2$.(complete later)
abstract-algebra group-theory order-theory
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
add a comment |
Definition: Given a left-orderable group $G$ with a left ordering $<$, a subgroup $H$ of $G$ is convex with respect to $<$ if for any elements $a<b in H$ and $a<c<b$ then $c in H$. A subgroup $Hsubset G$ is called relatively convex if there is an ordering in which $H$ is convex with respect to that ordering.
Prove: Suppose that $C$ is a subgroup of $G$, denote the set of left cosets ${gC}_{gin G}$ by $G/C$. The subgroup $C$ is relatively convex in $G$ if and only if there exists an ordering $prec$ of the cosets $G/C$ that is invariant under left multiplication by $G$, i.e. $gC prec hC$ implies $fgC prec fhC$ for all $f,g,h in G$.
Here is the question:
Here is problem 1.8 mentioned in the question above:
Here is my attempt:
Suppose $C$ is relatively convex, then there is an ordering $<$ on $G$ with respect to which, $C$ is convex. Then we define the ordering on the cosets of $C$ by $gC<hC$ if $g<h$. Now we need to show $<$ is left invariant and well defined. We first show that it is well defined: so suppose $g=g'c_1$, $h=h'c_2$, $g<h$ and $c_1<c_2$.(complete later)
abstract-algebra group-theory order-theory
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
Definition: Given a left-orderable group $G$ with a left ordering $<$, a subgroup $H$ of $G$ is convex with respect to $<$ if for any elements $a<b in H$ and $a<c<b$ then $c in H$. A subgroup $Hsubset G$ is called relatively convex if there is an ordering in which $H$ is convex with respect to that ordering.
Prove: Suppose that $C$ is a subgroup of $G$, denote the set of left cosets ${gC}_{gin G}$ by $G/C$. The subgroup $C$ is relatively convex in $G$ if and only if there exists an ordering $prec$ of the cosets $G/C$ that is invariant under left multiplication by $G$, i.e. $gC prec hC$ implies $fgC prec fhC$ for all $f,g,h in G$.
Here is the question:
Here is problem 1.8 mentioned in the question above:
Here is my attempt:
Suppose $C$ is relatively convex, then there is an ordering $<$ on $G$ with respect to which, $C$ is convex. Then we define the ordering on the cosets of $C$ by $gC<hC$ if $g<h$. Now we need to show $<$ is left invariant and well defined. We first show that it is well defined: so suppose $g=g'c_1$, $h=h'c_2$, $g<h$ and $c_1<c_2$.(complete later)
abstract-algebra group-theory order-theory
abstract-algebra group-theory order-theory
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
edited Nov 23 '18 at 13:12
mathnoob
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
asked Nov 11 '18 at 17:22
mathnoobmathnoob
1,792422
1,792422
add a comment |
add a comment |
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