Assuming this is the discrete Heisenberg group $H=H_3({mathbb Z})$, as in my comment above, then here is another way of looking at Mariano's answer (I think). Take any sequence of positive integers
$n_1 < n_2 < dots $ where $n_i vert n_{i+1}$ for all $i$, and put
$$ H_i = H_3(n_i{mathbb Z}) $$
(Mariano's answer corresponds to taking $n_i = 2^i$.)
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