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 $L$? 



Spanning set for Lattice generated by an orbit of the group.
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 $\pm 4\pm 1\neq \pm 1$. 


