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Feb 20 at 1:08 history edited LSpice CC BY-SA 4.0
Consistent TeX in title, while this is on the front page
Aug 22, 2011 at 13:19 vote accept Mustafa Gokhan Benli
Aug 19, 2011 at 3:27 answer added Ben Wieland timeline score: 7
Aug 19, 2011 at 3:22 answer added Ian Agol timeline score: 5
Aug 19, 2011 at 1:15 comment added Mustafa Gokhan Benli Thanks Agol and Richard. I think I can show that if $G\H$ is $\mathbb{Z}$ (maybe also if the quotient is free) and $H$ is finitely generated then the multiplier is always a direct product of f.g. abelian groups. SO one has to look for other type of examples.
Aug 19, 2011 at 0:45 comment added Autumn Kent The problem with my answer was that my subgroup $H$ isn't finitely generated (which is what Mustafa wants). I would expect some variant would do the trick, though.
Aug 18, 2011 at 19:19 history asked Mustafa Gokhan Benli CC BY-SA 3.0