Let $\mathfrak{g}$ be an n-dimensional, rational, nilpotent Lie algebra with simply connected that lie group $G$. It is stated in some papers that if $A$ is an automorphism of $\mathfrak{g}$ which is hyperbolic and in some basis an element of $SL(n,\mathbb{Z})$, then there is a $\mathbb{Z}$ Lie subalgebra (or a subring) $\Gamma_0$and some integer $m$ s.t $\Gamma= exp(\Gamma_0)$ is a subgroup in $G$ and $A^m(\Gamma) = \Gamma$ and therefore $A$ induces an Anosov diffeomorphism $f_A$ of $G\backslash \Gamma$.
I was wondering if the condition that $A$ is hyperbolic has any importance on the existence of the lattice $\Gamma_0$ satisfying the properties above. My expectation is that if $A$ is not hyperbolic then $f_A$ is simply not Anosov but neverthless some diffeomorphism of $G\backslash\Gamma$. My idea is that if in some basis $A$ is in $SL(n,\mathbb{Z})$ then it means that it preserves the vectorial integer lattice $\mathbb{Z}^n \subset \mathfrak{g}$ written with respect to that basis. I think then it only remains to show that for nilpotent, rational Lie algebra, there always exists a sublattice of $\mathbb{Z}^n$ of the form $\Gamma_0 = \{(q_1z_1,...,q_nz_n)\quad z_i \in \mathbb{Z}\}$ for some fixed rational $\{q_i\}$ s.t $exp(\Gamma_0)$ is a subgroup.Then I think it suffices just to take $m$ s.t $A^m(q_1,...,q_n) = (q_1,...q_n)$ producing the required Automorphism. So is such a thing always true or if any body knows a good refence for the proof about the first statement about automorphisms of nilmanifolds that would also suffice. Thanks
