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May 12, 2012 at 18:14 vote accept Mike Jury
May 12, 2012 at 11:32 answer added Mikael de la Salle timeline score: 6
May 12, 2012 at 7:13 comment added Mikael de la Salle I think that there is a rather simple iff condition for $\alpha(S)=1$~: if $S=\{g_1,g_2,\dots,g_n\}$, $\alpha(S)=1$ iff $g_1^{-1} g_2,\dots g_1^{-1} g_n$ generate a free abelian subgroup in the abelianization of $G$. I do not have time right now to write a proof, but if you are interested I can do it when I come back.
May 12, 2012 at 5:01 vote accept Mike Jury
May 12, 2012 at 18:14
May 12, 2012 at 3:35 history edited Mike Jury CC BY-SA 3.0
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May 12, 2012 at 3:07 answer added Nik Weaver timeline score: 4
May 12, 2012 at 1:37 answer added Nik Weaver timeline score: 2
May 12, 2012 at 1:24 history asked Mike Jury CC BY-SA 3.0