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Aug 13, 2013 at 19:14 vote accept Jiang
Aug 13, 2013 at 2:47 comment added Steve D @Jiang: For your group to be (virtually) nilpotent, $\sigma(a)$ must fix a point (otherwise the center of $G$ would be trivial). You can also check this is a sufficient condition (quotient by the fixed subgroup, and check it is (virtually) abelian).
Aug 11, 2013 at 22:17 comment added Jiang @Lee, especially, is the lemma 1 in Milnor's paper useful in our situation?
Aug 11, 2013 at 22:14 comment added Jiang @Lee, in your answer, you mentioned Milnor's paper, I checked it, but it is still not clear to me how to relate the nilpotentness of $G$ to the property of $\sigma(a)$, could you give more hints?
Aug 11, 2013 at 16:02 comment added Lee Mosher The theorem that $H_3$ and other (virtually) nilpotent groups have polynomial growth is a theorem of Milnor, with the exact degree of polynomial growth computed by Bass. Gromov's theorem is the converse: every group of polynomial growth is virtually nilpotent.
Aug 11, 2013 at 15:57 answer added Lee Mosher timeline score: 4
Aug 11, 2013 at 14:35 history asked Jiang CC BY-SA 3.0