Proof of group theory [closed]












-1














“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, Saad Nov 22 '18 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, Saad

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 '18 at 5:07










  • Finite index, I meant. I’m not sure it’s correct in English. Sorry.
    – saki
    Nov 20 '18 at 6:04
















-1














“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, Saad Nov 22 '18 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, Saad

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 '18 at 5:07










  • Finite index, I meant. I’m not sure it’s correct in English. Sorry.
    – saki
    Nov 20 '18 at 6:04














-1












-1








-1


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 10:18

























asked Nov 20 '18 at 4:06









saki

296




296




closed as off-topic by John Douma, Robert Z, Derek Holt, Shaun, Saad Nov 22 '18 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, Saad

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, Saad Nov 22 '18 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, Saad

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 '18 at 5:07










  • Finite index, I meant. I’m not sure it’s correct in English. Sorry.
    – saki
    Nov 20 '18 at 6:04














  • 1




    What's "limited index" mean?
    – Lord Shark the Unknown
    Nov 20 '18 at 5:07










  • Finite index, I meant. I’m not sure it’s correct in English. Sorry.
    – saki
    Nov 20 '18 at 6:04








1




1




What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 '18 at 5:07




What's "limited index" mean?
– Lord Shark the Unknown
Nov 20 '18 at 5:07












Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 '18 at 6:04




Finite index, I meant. I’m not sure it’s correct in English. Sorry.
– saki
Nov 20 '18 at 6:04










1 Answer
1






active

oldest

votes


















1














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 '18 at 12:39










  • Yes, that is what I mean.
    – Levent
    Nov 20 '18 at 15:31










  • I see, thank you very much!
    – saki
    Nov 20 '18 at 21:26


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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 '18 at 12:39










  • Yes, that is what I mean.
    – Levent
    Nov 20 '18 at 15:31










  • I see, thank you very much!
    – saki
    Nov 20 '18 at 21:26
















1














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 '18 at 12:39










  • Yes, that is what I mean.
    – Levent
    Nov 20 '18 at 15:31










  • I see, thank you very much!
    – saki
    Nov 20 '18 at 21:26














1












1








1






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 '18 at 11:51









Levent

2,688925




2,688925












  • 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 '18 at 12:39










  • Yes, that is what I mean.
    – Levent
    Nov 20 '18 at 15:31










  • I see, thank you very much!
    – saki
    Nov 20 '18 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 '18 at 12:39










  • Yes, that is what I mean.
    – Levent
    Nov 20 '18 at 15:31










  • I see, thank you very much!
    – saki
    Nov 20 '18 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 '18 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 '18 at 12:39












Yes, that is what I mean.
– Levent
Nov 20 '18 at 15:31




Yes, that is what I mean.
– Levent
Nov 20 '18 at 15:31












I see, thank you very much!
– saki
Nov 20 '18 at 21:26




I see, thank you very much!
– saki
Nov 20 '18 at 21:26



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