Re: [math-fun] Tiling puzzle
I wrote to Gene (inadvertently not sending it to all math-fun) about his solution to the tiling problem: << Yes, that's essentially what I had in mind -- nice solving. and then I added: << Now for Part B: Is there a way to solve this puzzle (for 2 sets) so that one of the sets is a subgroup of R ?
That is: Can R be partitioned into two congruent subsets that are each dense, one of which is a subgroup of R ? --Dan Even though kleptomaniacs can't help themselves, they do.
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
participants (2)
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Dan Asimov -
Fred W. Helenius