I have proven that every discrete subgroup of $\mathbb{R}^n$ is isomorphic to a $\mathbb{Z}^m$, but with no condition on $m$ (potentially, $m>n$). How do I prove that $m\leq n$?
I have tried doing it through an induction, which seems easiest but I can't really get the induction down (basis is easy).
On
I have also found an answer, more fitting to the method I used for solving the problem. Using the notations from my previous question, we have $P$ finite, so if we denote $l=\#(P)$ and $r=lcm(1,...,l)$ we have:
$$\forall x \in G, \exists (n_1,...,n_p) \in \mathbb{Z}^p, rx = n_1 a_1+...+n_p a_p$$
And therefore, $G$ can be injected in $\mathbb{Z}^n$, and is therefore isomorphic to a $\mathbb{Z}^m, m\leq n$ (see previous question and answer).
Suppose you have $m>n$. You may throw away some generators and assume that $m=n+1$. In this case you have $n+1$ generators of your group $v_1,\ldots,v_{n+1}$. They cannot be linearly independent over $\Bbb R$, but they are linearly independent over $\Bbb Q$. We can assume for some $r\le n$ that $v_1,\ldots,v_r$ are linearly independent over $\Bbb R$ but that $$v_{r+1}=a_1 v_1+\cdots +a_r v_r$$ with the $a_i\in\Bbb R$. Without loss we may assume $a_1\notin\Bbb Q$. For each $k\in \Bbb N$, $$\{ka_1\}v_1+\cdots+\{k a_r\}v_r\in G$$ where $\{x\}$ denotes the integer part of $x$. As all the $\{ka_1\}$ are distinct, these are infinitely many elements of $G$ in a bounded region, contradicting discreteness.