# Spanning set for Lattice generated by an orbit of the group.

For a vector spaces it always holds that any set of vectors spanning vector space $V$ has a subset of vectors which is a basis for $V$. While for lattices it is not true. For example consider one dimensional lattice spanned by $2,3$ then this lattice is $\mathbb{Z}$ and it is not spanned by any one of vectors. Simply stated my question is the following. Let $L$ be a lattice generated by an orbit of some vector $v\in V$ of dimension $k$. Does it always possible to find subset of the orbit which is a basis for lattice?

More formally: Let $G$ be a finite group. Let $\rho:G\rightarrow GL(V)$ be a representation such that $\rho(g)$ is an integer matrix for every $g$. Let $w\in V$ consider a lattice $$L= span_{\mathbb{Z}} ( {\rho(g)w: g\in G }).$$ Does it always possible to find $g_1,g_2,\ldots g_k$, where $k$ is the dimension of $L$ such that $( \rho(g_i)w)_{i=1}^k$ is a basis over $\mathbb{Z}$ for $L$?

• Your question is so terse that I can't easily tell if this is relevant or not, but it seems like it might be: blog.eqnets.com/2009/08/17/… – Steve Huntsman Mar 1 '12 at 18:33
• For me also the question seems out of focus. What exactly is the meaning here of "lattice", for instance? There is a large literature on integral representations of finite groups, to which it would be helpful to relate the question. – Jim Humphreys Mar 1 '12 at 19:34

## 1 Answer

It seems to me that the answer is no. For instance, let $\rho(G)$ consist of the $2\times 2$ signed permutation matrices, a group of order 8. Let $w=(2,1)$. The lattice $L$ is all of $\mathbb{Z}^2$ since for instance $(2,-1)+(-2,-1)+(1,2)= (1,0)$. But no two of $(\pm 1,\pm 2)$ and $(\pm 2,\pm 1)$ generate $\mathbb{Z}^2$ since the determinant of the matrix whose rows are any two of these eight vectors is either even or of the form $\pm 4\pm 1\neq \pm 1$.