Suppose we have an invertible matrix q in a finite subgroup $Q$ of $Gl(n,\mathbb Z)$, the group of all invertible integer matrices. Now I want to find all $x\; mod\; \mathbb Z^n$ for which

$(q+q^2+q^3+...+q^m).x = 0\quad mod\; \mathbb Z^n$

where $m$ is the order of $q$ in the finite subgroup $Q$ of $Gl(n,\mathbb Z)$ so that $q^m=1$. I tried using the Smith normal form so that

$(q+q^2+q^3+...+q^m) = U.D.V$

where $U,V$ in $Gl(n,\mathbb Z)$ and $D$ the Smith normal form, so we have to solve

$D.V.x=0\quad mod\; \mathbb Z^n$

Since $D.V$ is diagonal, $x$ must have rational components unless the diagonal element is zero. Now my question is, what is the maximal denominator of the components in $x$ ? So what is the maximal absolute value in $D.V$ ?I think this must be $m$, but I can't figure out why.

**Edit:**
Let me clarify why I expect x to be rational with an upper bound on the denominator. Suppose G is a subgroup of the Euclidean Group with isometries (t,q) as elements (t: translational part, q: linear part). The subgroup T which contains all isometries in G with trivial linear part is a normal subgroup of G. Suppose now that T can be identified with a $\mathbb Z$-lattice in $\mathbb R^n$, then G/T is isomorph with a finite subgroup Q of $GL(n,\mathbb Z)$. Crystallographers call G a space group and Q a point group.

There are only finite many conjugacy classes of finite subgroups in $GL(n,\mathbb Z)$, so there are only finite many point groups up to conjugacy in $GL(n,\mathbb Z)$. Now I want to understand why from this finite number of point groups, a finite number of (non-equivalent) space groups can be deduced. If we write G as the union of cosets of T

$G=\bigcup_{i=1}^{|Q|}(t_{qi},q_{i})T$

we see that (composition of two isometries and q belongs to exactly one coset)

$t_{q_1.q_2}=t_{q_1}+q_1.t_{q_2} \quad mod\ \mathbb Z^n$

So we know that $t_{q}$ is a real vector $0\leq t_{q}<1$. Using the previous property we also find that (m order of q)

$(t_{q},q)^{m}=(q^{1}\cdot t_{q}+\cdots+q^{m}\cdot t_{q},q^m)\in (0,id)T$

$\Leftrightarrow (q^{1}+\cdots+q^{m})\cdot t_{q}=0\quad mod\ \mathbb{Z}^{n}$

If an appropriate origin is chosen in Euclidean space, $t_{q}$ should be rational with maximal denominator $m$. Maybe investigating $(t_{q},q)^{m}$ is not the best way to find bounds on $t_{q}$?

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