Does there exist a finite-index subgroup of the real numbers under addition? O_o
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Anonymous2009-12-22 18:38
Hmmm, I'd say almost certainly yes.
If so, have Such a subgroup H in R.
Then R/H is a group of order n.
consider rH (coset), H=(rH)^n=r^nH and so r^n=nr is in H.
but notice that for all r, r/n is in R, and so by the previous argument n(r/n)=r is in H.
But than R is a subgroup of H. CONTRADICTION.
God I'm good.
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Anonymous2009-12-22 18:39
>>2
Notice I said almost certainly yes, then proved no. I'm even better then I first though
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Anonymous2009-12-22 23:45
>>2
R/H is an *additive* group of order n, so you can't say H = (rH)^n
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Anonymous2009-12-22 23:53
>>4
Yes you can. It's common (if misleading) notation, although I agree that using a+b instead of a*b and n*a instead of a^n makes more sense for additive groups. Note that he even wrote r^n=nr. The proof is correct despite the notation.
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Anonymous2009-12-23 0:01
Well, technically every group is a finite index subgroup of itself
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Anonymous2009-12-23 4:11
>>6 Well, technically...
nothing worthwhile has every come after that