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Apr 13, 2017 at 12:58 history edited CommunityBot
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Mar 14, 2012 at 0:49 history edited user6976 CC BY-SA 3.0
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Feb 10, 2011 at 1:36 comment added user6976 @Kevin: yes, see Andreas's answer below.
Feb 9, 2011 at 22:17 comment added Kevin Buzzard For the updated question, you may as well replace the subsets with the subgroups they generate, right? So an equivalent question is: "if a group $G$ is the union of finitely many subgroups, is one of the subgroups of finite index?".
Feb 9, 2011 at 22:15 answer added Andreas Thom timeline score: 13
Feb 9, 2011 at 21:51 history edited user6976 CC BY-SA 2.5
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Feb 9, 2011 at 20:35 vote accept CommunityBot moved from User.Id=6976 by developer User.Id=69903
Feb 9, 2011 at 20:35 comment added user6976 @Andreas: You are correct, one need to generate up to finite index.
Feb 9, 2011 at 20:01 comment added Andreas Thom To generate $G$ seems to be too much. I asked a modified form of this question as mathoverflow.net/questions/54921/…
Feb 9, 2011 at 19:22 answer added Stephen S timeline score: 33
Feb 9, 2011 at 18:04 comment added user6976 Example: if $G$ is the additive group of integers, then $S$ may consist of the first $2/\epsilon$ prime numbers (since every two primes generate $G$).
Feb 9, 2011 at 17:59 comment added user6976 $S$ depends on $\epsilon$ and $G$, so $|S|\gg 1/\epsilon$.
Feb 9, 2011 at 17:47 comment added ARupinski Just to clarify that I am reading this correctly, we are fixing $G$ and $\epsilon$ and trying to determine if such an $S$ exists for the pair $(G,\epsilon)$?
Feb 9, 2011 at 17:27 history asked user6976 CC BY-SA 2.5