Proof of group theory [closed]
up vote
-1
down vote
favorite
“Prove that the additive group of real numbers doesn’t have any proper subgroup with finite index.”
I want to know how to prove this.
group-theory
closed as off-topic by John Douma, Robert Z, Derek Holt, Shaun, user302797 Nov 22 at 3:05
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." – John Douma, Robert Z, Derek Holt, Shaun, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
“Prove that the additive group of real numbers doesn’t have any proper subgroup with finite index.”
I want to know how to prove this.
group-theory
closed as off-topic by John Douma, Robert Z, Derek Holt, Shaun, user302797 Nov 22 at 3:05
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." – John Douma, Robert Z, Derek Holt, Shaun, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
“Prove that the additive group of real numbers doesn’t have any proper subgroup with finite index.”
I want to know how to prove this.
group-theory
“Prove that the additive group of real numbers doesn’t have any proper subgroup with finite index.”
I want to know how to prove this.
group-theory
group-theory
edited Nov 20 at 10:18
asked Nov 20 at 4:06
saki
225
225
closed as off-topic by John Douma, Robert Z, Derek Holt, Shaun, user302797 Nov 22 at 3:05
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." – John Douma, Robert Z, Derek Holt, Shaun, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John Douma, Robert Z, Derek Holt, Shaun, user302797 Nov 22 at 3:05
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." – John Douma, Robert Z, Derek Holt, Shaun, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04
add a comment |
1
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04
1
1
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Assume that $Gleqmathbb{R}$ is a subgroup with $[mathbb{R}:G]=ninmathbb{Z}_{>0}$. Then for any $xinmathbb{R}$ we have $ncdot xin G$ (why?). In particular, for any $xinmathbb{R}$, we have $x = ncdot(frac{x}{n})in G$ so $mathbb{R}=G$.
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assume that $Gleqmathbb{R}$ is a subgroup with $[mathbb{R}:G]=ninmathbb{Z}_{>0}$. Then for any $xinmathbb{R}$ we have $ncdot xin G$ (why?). In particular, for any $xinmathbb{R}$, we have $x = ncdot(frac{x}{n})in G$ so $mathbb{R}=G$.
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
add a comment |
up vote
1
down vote
accepted
Assume that $Gleqmathbb{R}$ is a subgroup with $[mathbb{R}:G]=ninmathbb{Z}_{>0}$. Then for any $xinmathbb{R}$ we have $ncdot xin G$ (why?). In particular, for any $xinmathbb{R}$, we have $x = ncdot(frac{x}{n})in G$ so $mathbb{R}=G$.
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assume that $Gleqmathbb{R}$ is a subgroup with $[mathbb{R}:G]=ninmathbb{Z}_{>0}$. Then for any $xinmathbb{R}$ we have $ncdot xin G$ (why?). In particular, for any $xinmathbb{R}$, we have $x = ncdot(frac{x}{n})in G$ so $mathbb{R}=G$.
Assume that $Gleqmathbb{R}$ is a subgroup with $[mathbb{R}:G]=ninmathbb{Z}_{>0}$. Then for any $xinmathbb{R}$ we have $ncdot xin G$ (why?). In particular, for any $xinmathbb{R}$, we have $x = ncdot(frac{x}{n})in G$ so $mathbb{R}=G$.
answered Nov 20 at 11:51
Levent
3,386825
3,386825
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
add a comment |
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Thank you! You mean this? f:R→G(x→nx) for any x in R, there exists x/n in R s.t. x=f(x/n)=n(x/n) in G. Hence R=G.
– saki
Nov 20 at 12:39
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
Yes, that is what I mean.
– Levent
Nov 20 at 15:31
I see, thank you very much!
– saki
Nov 20 at 21:26
I see, thank you very much!
– saki
Nov 20 at 21:26
add a comment |
1
What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 at 5:07
Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 at 6:04