Show $C subset G$ is relatively convex iff $G/C$ is left-orderable.
up vote
0
down vote
favorite
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
add a comment |
up vote
0
down vote
favorite
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
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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
edited 2 days ago
asked Nov 11 at 17:22
mathnoob
89813
89813
add a comment |
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994154%2fshow-c-subset-g-is-relatively-convex-iff-g-c-is-left-orderable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown