Every Archimedean left-ordered group is abelian. [closed]
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What are the proofs of "Every Archimedean left-ordered group is abelian."? Here is one from the book:
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asked Nov 22 at 19:27
mathnoob
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closed as off-topic by Leucippus, max_zorn, Jean-Claude Arbaut, Davide Giraudo, amWhy Nov 22 at 21:52
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What are the proofs of "Every Archimedean left-ordered group is abelian."? Here is one from the book:
group-theory
asked Nov 22 at 19:27
mathnoob
1,744322
closed as off-topic by Leucippus, max_zorn, Jean-Claude Arbaut, Davide Giraudo, amWhy Nov 22 at 21:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, max_zorn, Jean-Claude Arbaut, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23
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up vote
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What are the proofs of "Every Archimedean left-ordered group is abelian."? Here is one from the book:
group-theory
asked Nov 22 at 19:27
mathnoob
1,744322
What are the proofs of "Every Archimedean left-ordered group is abelian."? Here is one from the book:
group-theory
group-theory
asked Nov 22 at 19:27
mathnoob
1,744322
asked Nov 22 at 19:27
mathnoob
1,744322
asked Nov 22 at 19:27
mathnoob
1,744322
asked Nov 22 at 19:27
mathnoob
1,744322
asked Nov 22 at 19:27
mathnoob
1,744322
1,744322
closed as off-topic by Leucippus, max_zorn, Jean-Claude Arbaut, Davide Giraudo, amWhy Nov 22 at 21:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, max_zorn, Jean-Claude Arbaut, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Leucippus, max_zorn, Jean-Claude Arbaut, Davide Giraudo, amWhy Nov 22 at 21:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, max_zorn, Jean-Claude Arbaut, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23
add a comment |
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23
add a comment |
1 Answer
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Let $G$ be an Archimedean ordered group.
Suppose there is a minimal positive element $g in G$. Then $G=langle g rangle$. Because pick $h in G-langle p rangle$, then there is $n in mathbb{N}$ such that $g^n leq h < g^{n+1}$. Then $ 1 leq g^{-n}h < g$. So that contradict minimality of $g$. So $G$ is cyclic and hence abelian.
Suppose there is no smallest positive element in $G$, then suppose $g,hin G$ do not commute, then assume W.L.O.G that $h,g,ghg^{-1}h^{-1}$ are all positive( the other cases can be proved analogously). Then $exists 1<x in G$ such that $1<x^2<ghg^{-1}h^{-1}$ So then $exists n,m in mathbb{N}$ such that $x^n<g<x^{n+1}$ and $x^n<h<x^{n+1}$. Which means $g^{-1}<x^{-n}$ and $h^{-1}<x^{-n}$. Multiplying out these inequality gives: $ghg^{-1}h^{-1} < x^2$ contradiction.
answered Nov 22 at 19:27
mathnoob
1,744322
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
add a comment |
1 Answer
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1 Answer
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active
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active
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active
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up vote
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Let $G$ be an Archimedean ordered group.
Suppose there is a minimal positive element $g in G$. Then $G=langle g rangle$. Because pick $h in G-langle p rangle$, then there is $n in mathbb{N}$ such that $g^n leq h < g^{n+1}$. Then $ 1 leq g^{-n}h < g$. So that contradict minimality of $g$. So $G$ is cyclic and hence abelian.
Suppose there is no smallest positive element in $G$, then suppose $g,hin G$ do not commute, then assume W.L.O.G that $h,g,ghg^{-1}h^{-1}$ are all positive( the other cases can be proved analogously). Then $exists 1<x in G$ such that $1<x^2<ghg^{-1}h^{-1}$ So then $exists n,m in mathbb{N}$ such that $x^n<g<x^{n+1}$ and $x^n<h<x^{n+1}$. Which means $g^{-1}<x^{-n}$ and $h^{-1}<x^{-n}$. Multiplying out these inequality gives: $ghg^{-1}h^{-1} < x^2$ contradiction.
answered Nov 22 at 19:27
mathnoob
1,744322
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
add a comment |
up vote
0
down vote
Let $G$ be an Archimedean ordered group.
Suppose there is a minimal positive element $g in G$. Then $G=langle g rangle$. Because pick $h in G-langle p rangle$, then there is $n in mathbb{N}$ such that $g^n leq h < g^{n+1}$. Then $ 1 leq g^{-n}h < g$. So that contradict minimality of $g$. So $G$ is cyclic and hence abelian.
Suppose there is no smallest positive element in $G$, then suppose $g,hin G$ do not commute, then assume W.L.O.G that $h,g,ghg^{-1}h^{-1}$ are all positive( the other cases can be proved analogously). Then $exists 1<x in G$ such that $1<x^2<ghg^{-1}h^{-1}$ So then $exists n,m in mathbb{N}$ such that $x^n<g<x^{n+1}$ and $x^n<h<x^{n+1}$. Which means $g^{-1}<x^{-n}$ and $h^{-1}<x^{-n}$. Multiplying out these inequality gives: $ghg^{-1}h^{-1} < x^2$ contradiction.
answered Nov 22 at 19:27
mathnoob
1,744322
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $G$ be an Archimedean ordered group.
Suppose there is a minimal positive element $g in G$. Then $G=langle g rangle$. Because pick $h in G-langle p rangle$, then there is $n in mathbb{N}$ such that $g^n leq h < g^{n+1}$. Then $ 1 leq g^{-n}h < g$. So that contradict minimality of $g$. So $G$ is cyclic and hence abelian.
Suppose there is no smallest positive element in $G$, then suppose $g,hin G$ do not commute, then assume W.L.O.G that $h,g,ghg^{-1}h^{-1}$ are all positive( the other cases can be proved analogously). Then $exists 1<x in G$ such that $1<x^2<ghg^{-1}h^{-1}$ So then $exists n,m in mathbb{N}$ such that $x^n<g<x^{n+1}$ and $x^n<h<x^{n+1}$. Which means $g^{-1}<x^{-n}$ and $h^{-1}<x^{-n}$. Multiplying out these inequality gives: $ghg^{-1}h^{-1} < x^2$ contradiction.
answered Nov 22 at 19:27
mathnoob
1,744322
Let $G$ be an Archimedean ordered group.
Suppose there is a minimal positive element $g in G$. Then $G=langle g rangle$. Because pick $h in G-langle p rangle$, then there is $n in mathbb{N}$ such that $g^n leq h < g^{n+1}$. Then $ 1 leq g^{-n}h < g$. So that contradict minimality of $g$. So $G$ is cyclic and hence abelian.
Suppose there is no smallest positive element in $G$, then suppose $g,hin G$ do not commute, then assume W.L.O.G that $h,g,ghg^{-1}h^{-1}$ are all positive( the other cases can be proved analogously). Then $exists 1<x in G$ such that $1<x^2<ghg^{-1}h^{-1}$ So then $exists n,m in mathbb{N}$ such that $x^n<g<x^{n+1}$ and $x^n<h<x^{n+1}$. Which means $g^{-1}<x^{-n}$ and $h^{-1}<x^{-n}$. Multiplying out these inequality gives: $ghg^{-1}h^{-1} < x^2$ contradiction.
answered Nov 22 at 19:27
mathnoob
1,744322
answered Nov 22 at 19:27
mathnoob
1,744322
answered Nov 22 at 19:27
mathnoob
1,744322
answered Nov 22 at 19:27
mathnoob
1,744322
1,744322
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
add a comment |
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
You can't multiply inequalities using only left-invariance.
– YCor
Nov 23 at 21:27
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Archimedean ordered group are bi-invariant.
– mathnoob
Nov 23 at 22:24
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
Yes, it's what I said in my comment to the question. It's a 15-line proof.
– YCor
Nov 23 at 22:34
add a comment |
See Lemma 2.2 in arxiv.org/abs/1511.05088. It states that every Archimedean left order is bi-invariant. Then it becomes very standard.
– YCor
Nov 23 at 21:29
How does bi invariant implies abelian?
– mathnoob
Nov 23 at 22:23