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 $t\_{q}$?



Finite subgroup of $Gl(n,\mathbb Z)$ and congruences
**Edit:** I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows: 


Let $k$ be a field, $q \in GL\_n(k)$ a matrix of finite exponent $m$ with char$(k) \nmid m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$: 



> 
> * $\lbrace x \in k^n \mid Ax \in M \rbrace = U + \frac{1}{m}(E \cap M)$
> * $U$ and $(1/m)(E \cap M)$ intersect in $0$ iff $0 \in M$, otherwise the intersection is empty
> * $A$ is diagonizable with diagonal $(m,...,m,0,...,0)$ where the number of m's equals $\dim E$
> 
> 
> 




---


*(Older formulation)* 


Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$. 


Set $A := 1+q+\dots + q^{m-1}$ and $X:= \lbrace x \in \mathbb{C}^n \mid A\cdot x \in \mathbb{Z}^n \rbrace$ and $L := E \cap \mathbb{Z}^n$. 


Then the following holds: 



> 
> $X = U \oplus \frac{1}{m}L$.
> 
> 
> 


*Proof*: Assume $\dim E = d$. Then $\dim U = \text{rank}(1-q) = n-d$. 


Since each $x \in E$ satisfies $Ax = mx$, $E$ contains eigenvectors from $A$ of the eigenvalue $m$. From $A \cdot (1-q) = 0$ it follows that $U$ consists of eigenvectors of $A$ of the eigenvalue $0$. Hence $E \cap U = 0$ and for dimensional reasons 
$$\mathbb{C}^n = U \oplus E.$$
Since $q$ has integral entries, it's possible to chosse a basis of $E$ in $\mathbb{Q}^n$ and by multiplying with a suitable integer it's also possible to choose a basis in $\mathbb{Z}^n$. Therefore $L = E \cap \mathbb{Z}^n$ is a lattice of rank $d$. Let $\lbrace e\_1, \dots, e\_d \rbrace$ be a basis of $L$. Let $x \in X$ and write 
$$x = u + \sum\_i \alpha\_i e\_i \text{ with } \alpha\_i \in \mathbb{C}.$$
Then $Ax = \sum\_i m\alpha\_i e\_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus\_i \mathbb{Z}e\_i$ and therefore $m\alpha\_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. *qed*. 


**Edit:** Also note that the image of $A$ is given by 
$$ Y := \lbrace Ax \mid x \in X \rbrace = L.$$


