13 Nov
2011
13 Nov
'11
12:41 p.m.
On 11/13/2011 1:00 PM, Dan Asimov wrote:
That is: Can R be partitioned into two congruent subsets that are each dense, one of which is a subgroup of R ?
No. The subgroup would have index two in R; this is impossible since R is a divisible group* and a divisible group cannot have a proper finite-index subgroup. A quick proof can be found in Pete L. Clark's answer on this page: http://mathoverflow.net/questions/59167/profinite-completion *An abelian group G is called divisible if, for every g in G and every positive integer n, there exists an x in G such that nx = g. -- Fred W. Helenius fredh@ix.netcom.com