Skip to main content
5 events
when toggle format what by license comment
Jan 28, 2010 at 20:31 answer added Rishi Vyas timeline score: 7
Jan 28, 2010 at 16:09 comment added Anweshi What if one requires in addition that $G$ has no torsion?
Jan 28, 2010 at 16:01 comment added DoubtingThomas Thanks. Now perhaps it's time to close the question?
Jan 28, 2010 at 15:57 comment added Emerton Maybe maximal subguroup means maximal proper subgroup and Q-algebra means Q-vector space? With these interpretations the answer is no: Q/Z has no maximal proper subgroup, and is not a Q-vector space.
Jan 28, 2010 at 15:43 history asked DoubtingThomas CC BY-SA 2.5