discrete subgroups of euclidean space I'm trying to prove this proposition:
http://groupprops.subwiki.org/wiki/Every_discrete_subgroup_of_Euclidean_space_is_free_Abelian_on_a_linearly_independent_set
That every discrete subgroup of Euclidean space is free Abelian with rank at most n, but I don't know how to approach it, even with the "used facts" section.
Any help please? 
 A: Let $h\in H$ since $h+h+...+h\in H$ this means that $h\mathbb{Z}\subset H$. Now assume there are at most $k\leq n$ linearly independant (over the reals) $h_i\in H$. This means that $H$ contains the lattice generated by them which is a discrete abelien subgroup of $\mathbb{R}^n$ and has rank $k$.
Since we assumed $H$ has only $k$ linearly independat vectors that means that $H$ spans a $k$-dimensional subspace of $\mathbb{R}^n$ which for simplicity we shall denote by $\mathbb{R}^k$. 
So we have a full lattice $L \subset H \subset \mathbb{R}^k$.
We need to show that $H$ is also a lattice. This would mean it has $k$ generators and therefore rank $k$.
That would need to show that the index $|H:L|$ is finite.
Now let is notice that we can represent any vector $h\in H$ as $h=l+p$ for $l\in L$ and $p=(a_1h_1,...,a_kh_k)$ for $a_i\in [0,1)$ (the set $P$ of such points $p$ is called the fundamental parallelepiped of a lattice).
Now since $H$ is a group $p\in H$.
$P$ however is bounded and $H$ is discrete therefore there can only be finitely many elements $p\in P$ which are also in $H$.
This means that $\forall p\in H \exists i_p: i_pp\in H$. Let $j$ the least common multiple.
This means that $H$ is contained in the lattice generated by $\frac{h_i}{j}$.
Hence $H$ is a lattice and has rank $k$.
