Proof of group theory [closed]











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“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







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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

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    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















up vote
-1
down vote

favorite
2












“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







share|cite|improve this 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













up vote
-1
down vote

favorite
2









up vote
-1
down vote

favorite
2






2





“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







share|cite|improve this question















“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






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share|cite|improve this question













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share|cite|improve this question








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














  • 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










1 Answer
1






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1
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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 $mathbb{R}=G$.






share|cite|improve this answer





















  • 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


















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 $mathbb{R}=G$.






share|cite|improve this answer





















  • 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















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 $mathbb{R}=G$.






share|cite|improve this answer





















  • 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













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 $mathbb{R}=G$.






share|cite|improve this answer












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 $mathbb{R}=G$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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



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