Subgroups of $Bbb R^times$

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4
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1. What are the open subgroups of $Bbb R^times$ ?


2. What are the closed subgroups of $Bbb R^times$ ?


3. What are the finite index subgroups of $Bbb R^times$ ?

My thoughts:

1. Since $Bbb R_>0$ is a connected Lie group, any open neighborhood of $1$ generates it as a group. So only $Bbb R_>0$ and $Bbb R^times$ are the open subgroups of $Bbb R^times$. Is it correct?




2. Here I don't really know.




3. If $H subset Bbb R_>0$ has finite index in $Bbb R^times$ then it is $Bbb R_>0$ by this answer – because $ Bbb R_>0$ is divisible. If $H subset Bbb R^times$ is a subgroup of finite index not contained in $Bbb R_>0$, do we have $H = Bbb R^times$ ? Then we would have only 2 subgroups of finite index.

abelian-groups topological-groups

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asked Jul 15 at 9:51

Alphonse

1,767622

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For 2., do you know the closed subgroups of $Bbb R^+$?
  – Lord Shark the Unknown
  Jul 15 at 9:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
  – Alphonse
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is $n$ there?
  – Lord Shark the Unknown
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
  – Lord Shark the Unknown
  Jul 15 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
  – Alphonse
  Jul 15 at 10:02
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

1. What are the open subgroups of $Bbb R^times$ ?


2. What are the closed subgroups of $Bbb R^times$ ?


3. What are the finite index subgroups of $Bbb R^times$ ?

My thoughts:

1. Since $Bbb R_>0$ is a connected Lie group, any open neighborhood of $1$ generates it as a group. So only $Bbb R_>0$ and $Bbb R^times$ are the open subgroups of $Bbb R^times$. Is it correct?




2. Here I don't really know.




3. If $H subset Bbb R_>0$ has finite index in $Bbb R^times$ then it is $Bbb R_>0$ by this answer – because $ Bbb R_>0$ is divisible. If $H subset Bbb R^times$ is a subgroup of finite index not contained in $Bbb R_>0$, do we have $H = Bbb R^times$ ? Then we would have only 2 subgroups of finite index.

abelian-groups topological-groups

share|cite|improve this question

asked Jul 15 at 9:51

Alphonse

1,767622

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For 2., do you know the closed subgroups of $Bbb R^+$?
  – Lord Shark the Unknown
  Jul 15 at 9:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
  – Alphonse
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is $n$ there?
  – Lord Shark the Unknown
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
  – Lord Shark the Unknown
  Jul 15 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
  – Alphonse
  Jul 15 at 10:02
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

up vote
4
down vote

favorite

1. What are the open subgroups of $Bbb R^times$ ?


2. What are the closed subgroups of $Bbb R^times$ ?


3. What are the finite index subgroups of $Bbb R^times$ ?

My thoughts:

1. Since $Bbb R_>0$ is a connected Lie group, any open neighborhood of $1$ generates it as a group. So only $Bbb R_>0$ and $Bbb R^times$ are the open subgroups of $Bbb R^times$. Is it correct?




2. Here I don't really know.




3. If $H subset Bbb R_>0$ has finite index in $Bbb R^times$ then it is $Bbb R_>0$ by this answer – because $ Bbb R_>0$ is divisible. If $H subset Bbb R^times$ is a subgroup of finite index not contained in $Bbb R_>0$, do we have $H = Bbb R^times$ ? Then we would have only 2 subgroups of finite index.

abelian-groups topological-groups

share|cite|improve this question

asked Jul 15 at 9:51

Alphonse

1,767622

1. What are the open subgroups of $Bbb R^times$ ?


2. What are the closed subgroups of $Bbb R^times$ ?


3. What are the finite index subgroups of $Bbb R^times$ ?

My thoughts:

1. Since $Bbb R_>0$ is a connected Lie group, any open neighborhood of $1$ generates it as a group. So only $Bbb R_>0$ and $Bbb R^times$ are the open subgroups of $Bbb R^times$. Is it correct?




2. Here I don't really know.




3. If $H subset Bbb R_>0$ has finite index in $Bbb R^times$ then it is $Bbb R_>0$ by this answer – because $ Bbb R_>0$ is divisible. If $H subset Bbb R^times$ is a subgroup of finite index not contained in $Bbb R_>0$, do we have $H = Bbb R^times$ ? Then we would have only 2 subgroups of finite index.

abelian-groups topological-groups

share|cite|improve this question

asked Jul 15 at 9:51

Alphonse

1,767622

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 9:51

Alphonse

1,767622

asked Jul 15 at 9:51

Alphonse

1,767622

asked Jul 15 at 9:51

Alphonse

1,767622

1,767622

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For 2., do you know the closed subgroups of $Bbb R^+$?
  – Lord Shark the Unknown
  Jul 15 at 9:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
  – Alphonse
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is $n$ there?
  – Lord Shark the Unknown
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
  – Lord Shark the Unknown
  Jul 15 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
  – Alphonse
  Jul 15 at 10:02
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For 2., do you know the closed subgroups of $Bbb R^+$?
  – Lord Shark the Unknown
  Jul 15 at 9:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
  – Alphonse
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is $n$ there?
  – Lord Shark the Unknown
  Jul 15 at 9:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
  – Lord Shark the Unknown
  Jul 15 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
  – Alphonse
  Jul 15 at 10:02
  

  
  
  
  

4

4

For 2., do you know the closed subgroups of $Bbb R^+$?
– Lord Shark the Unknown
Jul 15 at 9:52

For 2., do you know the closed subgroups of $Bbb R^+$?
– Lord Shark the Unknown
Jul 15 at 9:52

2

2

@LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
– Alphonse
Jul 15 at 9:54

@LordSharktheUnknown Ah yes, $Bbb R^times = pm 1 times (Bbb R,+)$. I know $(n Bbb Z, +)$ as closed subgroups of $(Bbb R,+)$. Are there other?
– Alphonse
Jul 15 at 9:54

2

2

What is $n$ there?
– Lord Shark the Unknown
Jul 15 at 9:54

What is $n$ there?
– Lord Shark the Unknown
Jul 15 at 9:54

2

2

A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
– Lord Shark the Unknown
Jul 15 at 10:00

A proper closed subgroup of $Bbb R$ would meet some interval $(-epsilon,epsilon)$ only at $0$. Could a non-cyclic subgroup do that?
– Lord Shark the Unknown
Jul 15 at 10:00

2

2

@LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
– Alphonse
Jul 15 at 10:02

@LordSharktheUnknown : ok… any non-cyclic subgroup of $Bbb R$ is dense :-). Thanks!
– Alphonse
Jul 15 at 10:02

 | 
show 2 more comments

1 Answer
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2
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Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2.

The key observation is that most the closed subgroup of $mathbbR^times$ generated by only two elements is at least $mathbbR_>0$ unless something very special happens.

Lemma. Let $x, y in mathbbR_>0$ and put $A = x^n y^m : n,m in mathbbZ$. Then either (1) $A = z^n : n in mathbbZ $ for some $z in mathbbR_>0$, or (2) $A$ is dense in $mathbbR_>0$.

Proof. Apply Kronecker's theorem to $log(x)/log(y)$.

Suppose that $G < mathbbR^times$ is a closed subgroup, and let $H = G cap mathbbR_>0$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H cup xH$ where $x in mathbbR_<0$ and $x^2 in H$. So, it's enough to characterise all possible $H$.

If $H$ is the trivial group, we are done, so suppose there is some $x in H setminus 1$. Likewise, if $H = mathbbR_>0$ we are done, so suppose that $H$ is not dense in $mathbbR_>0$. In particular, there are only finitely many positive integers $n$ such that $x^1/n in H$; let $t = x^1/n$ where $n$ is largest possible. For any $y in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = t^n : n in mathbbZ $ (the opposite inclusion is clear).

Conversely, for any $t > 0$, $H = t^n : n in mathbbZ $ is a closed subgroup of $mathbbR_>0$ for any $t geq 0$. Hence, the closed subgroups of $mathbbR^times$ are: $mathbbR^times$, $mathbbR_>0$, $t^n : n in mathbbZ $ for $t > 0$, $pm t^n : n in mathbbZ $ for $t > 0$, and $(-t)^n : n in mathbbZ $ for $t > 0$.

Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(mathbbR,+)$, but since the problems are very similar I avoid that assumption :-)

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2.

The key observation is that most the closed subgroup of $mathbbR^times$ generated by only two elements is at least $mathbbR_>0$ unless something very special happens.

Lemma. Let $x, y in mathbbR_>0$ and put $A = x^n y^m : n,m in mathbbZ$. Then either (1) $A = z^n : n in mathbbZ $ for some $z in mathbbR_>0$, or (2) $A$ is dense in $mathbbR_>0$.

Proof. Apply Kronecker's theorem to $log(x)/log(y)$.

Suppose that $G < mathbbR^times$ is a closed subgroup, and let $H = G cap mathbbR_>0$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H cup xH$ where $x in mathbbR_<0$ and $x^2 in H$. So, it's enough to characterise all possible $H$.

If $H$ is the trivial group, we are done, so suppose there is some $x in H setminus 1$. Likewise, if $H = mathbbR_>0$ we are done, so suppose that $H$ is not dense in $mathbbR_>0$. In particular, there are only finitely many positive integers $n$ such that $x^1/n in H$; let $t = x^1/n$ where $n$ is largest possible. For any $y in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = t^n : n in mathbbZ $ (the opposite inclusion is clear).

Conversely, for any $t > 0$, $H = t^n : n in mathbbZ $ is a closed subgroup of $mathbbR_>0$ for any $t geq 0$. Hence, the closed subgroups of $mathbbR^times$ are: $mathbbR^times$, $mathbbR_>0$, $t^n : n in mathbbZ $ for $t > 0$, $pm t^n : n in mathbbZ $ for $t > 0$, and $(-t)^n : n in mathbbZ $ for $t > 0$.

Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(mathbbR,+)$, but since the problems are very similar I avoid that assumption :-)

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

add a comment | 

up vote
2
down vote

Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2.

The key observation is that most the closed subgroup of $mathbbR^times$ generated by only two elements is at least $mathbbR_>0$ unless something very special happens.

Lemma. Let $x, y in mathbbR_>0$ and put $A = x^n y^m : n,m in mathbbZ$. Then either (1) $A = z^n : n in mathbbZ $ for some $z in mathbbR_>0$, or (2) $A$ is dense in $mathbbR_>0$.

Proof. Apply Kronecker's theorem to $log(x)/log(y)$.

Suppose that $G < mathbbR^times$ is a closed subgroup, and let $H = G cap mathbbR_>0$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H cup xH$ where $x in mathbbR_<0$ and $x^2 in H$. So, it's enough to characterise all possible $H$.

If $H$ is the trivial group, we are done, so suppose there is some $x in H setminus 1$. Likewise, if $H = mathbbR_>0$ we are done, so suppose that $H$ is not dense in $mathbbR_>0$. In particular, there are only finitely many positive integers $n$ such that $x^1/n in H$; let $t = x^1/n$ where $n$ is largest possible. For any $y in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = t^n : n in mathbbZ $ (the opposite inclusion is clear).

Conversely, for any $t > 0$, $H = t^n : n in mathbbZ $ is a closed subgroup of $mathbbR_>0$ for any $t geq 0$. Hence, the closed subgroups of $mathbbR^times$ are: $mathbbR^times$, $mathbbR_>0$, $t^n : n in mathbbZ $ for $t > 0$, $pm t^n : n in mathbbZ $ for $t > 0$, and $(-t)^n : n in mathbbZ $ for $t > 0$.

Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(mathbbR,+)$, but since the problems are very similar I avoid that assumption :-)

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

add a comment | 

up vote
2
down vote

up vote
2
down vote

Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2.

The key observation is that most the closed subgroup of $mathbbR^times$ generated by only two elements is at least $mathbbR_>0$ unless something very special happens.

Lemma. Let $x, y in mathbbR_>0$ and put $A = x^n y^m : n,m in mathbbZ$. Then either (1) $A = z^n : n in mathbbZ $ for some $z in mathbbR_>0$, or (2) $A$ is dense in $mathbbR_>0$.

Proof. Apply Kronecker's theorem to $log(x)/log(y)$.

Suppose that $G < mathbbR^times$ is a closed subgroup, and let $H = G cap mathbbR_>0$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H cup xH$ where $x in mathbbR_<0$ and $x^2 in H$. So, it's enough to characterise all possible $H$.

If $H$ is the trivial group, we are done, so suppose there is some $x in H setminus 1$. Likewise, if $H = mathbbR_>0$ we are done, so suppose that $H$ is not dense in $mathbbR_>0$. In particular, there are only finitely many positive integers $n$ such that $x^1/n in H$; let $t = x^1/n$ where $n$ is largest possible. For any $y in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = t^n : n in mathbbZ $ (the opposite inclusion is clear).

Conversely, for any $t > 0$, $H = t^n : n in mathbbZ $ is a closed subgroup of $mathbbR_>0$ for any $t geq 0$. Hence, the closed subgroups of $mathbbR^times$ are: $mathbbR^times$, $mathbbR_>0$, $t^n : n in mathbbZ $ for $t > 0$, $pm t^n : n in mathbbZ $ for $t > 0$, and $(-t)^n : n in mathbbZ $ for $t > 0$.

Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(mathbbR,+)$, but since the problems are very similar I avoid that assumption :-)

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2.

The key observation is that most the closed subgroup of $mathbbR^times$ generated by only two elements is at least $mathbbR_>0$ unless something very special happens.

Lemma. Let $x, y in mathbbR_>0$ and put $A = x^n y^m : n,m in mathbbZ$. Then either (1) $A = z^n : n in mathbbZ $ for some $z in mathbbR_>0$, or (2) $A$ is dense in $mathbbR_>0$.

Proof. Apply Kronecker's theorem to $log(x)/log(y)$.

Suppose that $G < mathbbR^times$ is a closed subgroup, and let $H = G cap mathbbR_>0$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H cup xH$ where $x in mathbbR_<0$ and $x^2 in H$. So, it's enough to characterise all possible $H$.

If $H$ is the trivial group, we are done, so suppose there is some $x in H setminus 1$. Likewise, if $H = mathbbR_>0$ we are done, so suppose that $H$ is not dense in $mathbbR_>0$. In particular, there are only finitely many positive integers $n$ such that $x^1/n in H$; let $t = x^1/n$ where $n$ is largest possible. For any $y in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = t^n : n in mathbbZ $ (the opposite inclusion is clear).

Conversely, for any $t > 0$, $H = t^n : n in mathbbZ $ is a closed subgroup of $mathbbR_>0$ for any $t geq 0$. Hence, the closed subgroups of $mathbbR^times$ are: $mathbbR^times$, $mathbbR_>0$, $t^n : n in mathbbZ $ for $t > 0$, $pm t^n : n in mathbbZ $ for $t > 0$, and $(-t)^n : n in mathbbZ $ for $t > 0$.

Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(mathbbR,+)$, but since the problems are very similar I avoid that assumption :-)

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

answered Jul 15 at 12:09

Jakub Konieczny

9,21511759

9,21511759

add a comment | 

add a comment | 

 

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