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Totally ordered abelian group with a unique “ isolated subgroupâ€
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Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,bin G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and for every $a,b,cin G$, $a<b implies a+c<b+c$. Let us call a subgroup $H$ of a totally ordered abelian group $G$ to be isolated if $Hne G$ and $ain H, -a<a implies bin H, forall -a<b<a$.
My question is: If $(G,+,<)$ is a totally ordered abelian group with a unique isolated subgroup i.e. the trivial subgroup $e$ is the only isolated subgroup, then is it true that there is an order preserving isomorphism between $G$ and a subgroup of $(mathbb R,+)$ with the usual order inherited from real line ?
real-analysis abelian-groups ordered-groups
asked Jul 31 at 16:34
user521337
606
add a comment |Â
up vote
1
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Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,bin G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and for every $a,b,cin G$, $a<b implies a+c<b+c$. Let us call a subgroup $H$ of a totally ordered abelian group $G$ to be isolated if $Hne G$ and $ain H, -a<a implies bin H, forall -a<b<a$.
My question is: If $(G,+,<)$ is a totally ordered abelian group with a unique isolated subgroup i.e. the trivial subgroup $e$ is the only isolated subgroup, then is it true that there is an order preserving isomorphism between $G$ and a subgroup of $(mathbb R,+)$ with the usual order inherited from real line ?
real-analysis abelian-groups ordered-groups
asked Jul 31 at 16:34
user521337
606
Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,bin G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and for every $a,b,cin G$, $a<b implies a+c<b+c$. Let us call a subgroup $H$ of a totally ordered abelian group $G$ to be isolated if $Hne G$ and $ain H, -a<a implies bin H, forall -a<b<a$.
My question is: If $(G,+,<)$ is a totally ordered abelian group with a unique isolated subgroup i.e. the trivial subgroup $e$ is the only isolated subgroup, then is it true that there is an order preserving isomorphism between $G$ and a subgroup of $(mathbb R,+)$ with the usual order inherited from real line ?
real-analysis abelian-groups ordered-groups
asked Jul 31 at 16:34
user521337
606
Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,bin G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and for every $a,b,cin G$, $a<b implies a+c<b+c$. Let us call a subgroup $H$ of a totally ordered abelian group $G$ to be isolated if $Hne G$ and $ain H, -a<a implies bin H, forall -a<b<a$.
My question is: If $(G,+,<)$ is a totally ordered abelian group with a unique isolated subgroup i.e. the trivial subgroup $e$ is the only isolated subgroup, then is it true that there is an order preserving isomorphism between $G$ and a subgroup of $(mathbb R,+)$ with the usual order inherited from real line ?
real-analysis abelian-groups ordered-groups
asked Jul 31 at 16:34
user521337
606
edited Aug 2 at 0:21
asked Jul 31 at 16:34
user521337
606
asked Jul 31 at 16:34
user521337
606
asked Jul 31 at 16:34
user521337
606
606
Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38
add a comment |Â
Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38
Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38
add a comment |Â
1 Answer
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Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,bin G$, if $a>0$, then there is a $ninmathbbN$ such that $nageq b$).
Let $a>0$ be an element of $G$. Let $H=xin G:exists ninmathbbN:-na<x<na$. Then $-a<0<a$, so $0in H$. For $x,yin H$, there are $m,ninmathbbN$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-yin H$. Also, for $x,yin G$, if $0leq yleq x$ and $xin H$ then there is $ninmathbbN$ such that $0leq x<na$, so $0leq yleq x<na$, so $yin H$. Therefore, either $H$ is an isolated subgroup (then $H=0$) or $H=G$, but $-2a<a<2a$, so $ain H$, so $Hneq0$, so $H=G$. Therefore, $G$ is Archimedean.
answered Aug 2 at 9:51
Daniel Kawai
689
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,bin G$, if $a>0$, then there is a $ninmathbbN$ such that $nageq b$).
Let $a>0$ be an element of $G$. Let $H=xin G:exists ninmathbbN:-na<x<na$. Then $-a<0<a$, so $0in H$. For $x,yin H$, there are $m,ninmathbbN$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-yin H$. Also, for $x,yin G$, if $0leq yleq x$ and $xin H$ then there is $ninmathbbN$ such that $0leq x<na$, so $0leq yleq x<na$, so $yin H$. Therefore, either $H$ is an isolated subgroup (then $H=0$) or $H=G$, but $-2a<a<2a$, so $ain H$, so $Hneq0$, so $H=G$. Therefore, $G$ is Archimedean.
answered Aug 2 at 9:51
Daniel Kawai
689
add a comment |Â
up vote
0
down vote
accepted
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,bin G$, if $a>0$, then there is a $ninmathbbN$ such that $nageq b$).
Let $a>0$ be an element of $G$. Let $H=xin G:exists ninmathbbN:-na<x<na$. Then $-a<0<a$, so $0in H$. For $x,yin H$, there are $m,ninmathbbN$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-yin H$. Also, for $x,yin G$, if $0leq yleq x$ and $xin H$ then there is $ninmathbbN$ such that $0leq x<na$, so $0leq yleq x<na$, so $yin H$. Therefore, either $H$ is an isolated subgroup (then $H=0$) or $H=G$, but $-2a<a<2a$, so $ain H$, so $Hneq0$, so $H=G$. Therefore, $G$ is Archimedean.
answered Aug 2 at 9:51
Daniel Kawai
689
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,bin G$, if $a>0$, then there is a $ninmathbbN$ such that $nageq b$).
Let $a>0$ be an element of $G$. Let $H=xin G:exists ninmathbbN:-na<x<na$. Then $-a<0<a$, so $0in H$. For $x,yin H$, there are $m,ninmathbbN$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-yin H$. Also, for $x,yin G$, if $0leq yleq x$ and $xin H$ then there is $ninmathbbN$ such that $0leq x<na$, so $0leq yleq x<na$, so $yin H$. Therefore, either $H$ is an isolated subgroup (then $H=0$) or $H=G$, but $-2a<a<2a$, so $ain H$, so $Hneq0$, so $H=G$. Therefore, $G$ is Archimedean.
answered Aug 2 at 9:51
Daniel Kawai
689
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,bin G$, if $a>0$, then there is a $ninmathbbN$ such that $nageq b$).
Let $a>0$ be an element of $G$. Let $H=xin G:exists ninmathbbN:-na<x<na$. Then $-a<0<a$, so $0in H$. For $x,yin H$, there are $m,ninmathbbN$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-yin H$. Also, for $x,yin G$, if $0leq yleq x$ and $xin H$ then there is $ninmathbbN$ such that $0leq x<na$, so $0leq yleq x<na$, so $yin H$. Therefore, either $H$ is an isolated subgroup (then $H=0$) or $H=G$, but $-2a<a<2a$, so $ain H$, so $Hneq0$, so $H=G$. Therefore, $G$ is Archimedean.
answered Aug 2 at 9:51
Daniel Kawai
689
answered Aug 2 at 9:51
Daniel Kawai
689
answered Aug 2 at 9:51
Daniel Kawai
689
answered Aug 2 at 9:51
Daniel Kawai
689
689
add a comment |Â
add a comment |Â
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Another name for isolated subgroup is convex subgroup.
– Daniel Kawai
Aug 1 at 23:25
@DanielKawai : ok ... so ?
– user521337
Aug 2 at 0:21
YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0in G$ with $a_0>e.$ For $e<xin G$ and $nin Bbb N$ let $f(x,n)in Bbb N$ such that $nxleq a_0f(x,n)<(n+1)x.$ Prove that $psi (x)=lim_nto inftyn^-1f(x,n)$ exists. Let $psi (y)=-psi (-y)$ for $y<e$ and $psi (e)=0$. Now prove that $psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about.
– DanielWainfleet
Aug 2 at 1:33
The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<bimplies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian.
– DanielWainfleet
Aug 2 at 1:38