I am trying to prove that every closed discrete subgroup of $\Bbb{R}^n$ under addition is a free abelian group of finite rank. I have tried to do this by induction on the dimension $n$.
Base case: We claim that a subgroup $H \subset \Bbb{R}$ has a least positive element $\alpha$. The completeness axiom tells us that there is a least positive real number $\alpha$ among all positive reals in $H$, and being closed implies that $\alpha \in H$.
Now assume the inductive hypothesis and consider a closed, discrete subgroup $H$ of $\Bbb{R}^n$. Choose some element $x \in H$ of least positive distance to the origin. This can be done because otherwise we get a contradiction like the above. Now let $L$ be the subspace spanned by $x$ and write $\Bbb{R}^n = L \oplus W$ for some complement $W$. Consider the projection (as a linear map) $p : \Bbb{R}^n \to W$. Now this is also a group homomorphism and so we have
$\textrm{Im} (H) \subset W$
being a subgroup and so by the induction hypothesis is isomorphic to $\Bbb{Z}^l$ for some $1 \leq l \leq n-1$. However from here I am having some trouble concluding that $H$ itself must be isomorphic to $\Bbb{Z}^l \oplus \Bbb{Z}x$. If I know the existence of a map $l : \textrm{Im}(H) \to H$ such that
$p \circ l = \textrm{id}_{\textrm{Im}(H)}$
then by the splitting lemma I can conclude my problem. However, I don't have such a map so how do I conclude the problem? Please do not post complete solutions.
Thanks.
Edit: Perhaps I should add some context. I am trying to conclude that the kernel of the exponential map $\exp : \mathfrak{g} \to G$ where $G$ is a connected abelian matrix Lie group is a free abelian group of finite rank. We know that the assumptions on our matrix Lie group mean that $\exp$ is a group homomorphism, so that $\ker \exp$ is a subgroup of the Lie algebra $\mathfrak{g}$.
Edit: The problem is reduced to showing that the image of $H$ under the projection is indeed discrete, because otherwise we cannot apply the inductive hypothesis.