Subgroups of a vector space I would like to have an overview of how a subgroup of a vector space over $\mathbb R$ of dimension $n$ can look like.
Is there a complete classification available? I know that there are for examples the linear subspaces,
subgroups which are ismorphic to $\mathbb Z^k$ or $\mathbb  Q^k$ (with $k\leq n$)
and probably many more. 
What if I impose some extra condition? For example, I know that the only discrete subgroups
are those isomorphic to $\mathbb Z^k$.
So what happens if I ask for local compactness? (I know this rules out the subgroups isomorphic to $\mathbb  Q^k$).  Or for the Lie subgroups?
Are in this case the linear subspaces and the discrete subgroups the only candidates?  
 A: Every vector space over a field of positive characteristic $p$ is in particular a vector space over $\mathbb{F}_p$. Any subgroup of such a vector space is a subspace (exercise), and conversely. Assuming the axiom of choice, any such subspace is a direct sum of copies of $\mathbb{F}_p$. 
Every vector space over a field of characteristic zero is in particular a vector space over $\mathbb{Q}$. Assuming the axiom of choice, it is in fact a direct sum of copies of $\mathbb{Q}$. The subgroups of a direct sum must be direct sums of subgroups, so your question reduces to the classification of subgroups of $\mathbb{Q}$. 
Let $H$ be a subgroup of a $\mathbb{Q}$-vector space $V$. By restricting our attention to the subspace spanned by $H$, we may WLOG assume that $H$ spans $V$ as a vector space. Take a basis $e_i, i \in I$ of $V$ consisting of elements of $H$ and consider the short exact sequence
$$0 \to \bigoplus_i \mathbb{Z} e_i \to V \cong \bigoplus_i \mathbb{Q} e_i \to \bigoplus_i \mathbb{Q}/\mathbb{Z} \to 0.$$
Since $e_i \in H$ for all $i$, it's not hard to see that $H$ is necessarily the preimage of its image in $\bigoplus \mathbb{Q}/\mathbb{Z}$, so our problem reduces to the problem of classifying subgroups of $M = \bigoplus_i \mathbb{Q}/\mathbb{Z}$. By partial fraction decomposition (which makes sense for rational numbers just as well as rational functions even if nobody's ever told you this!), $M$ is the direct sum of its Sylow $p$-subgroups
$$M_p \cong \bigoplus_i \mathbb{Z}(p^{\infty})$$
where $\mathbb{Z}(p^{\infty})$ is the Prüfer $p$-group (I really don't like this notation but it appears to be standard). 
Proposition: Let $H$ be a subgroup of $M$, and let $H_p$ be its image in $M_p$ (regarded as a subgroup of $M$). Then $H = \bigoplus_p H_p$.
Proof. Let $h \in H$ be an element. By multiplying $h$ by appropriate powers of every prime not equal to $p$ which occurs in the denominators of the components of $h$, we can find an element which has the same image as $h$ in $M_p$ but which has zero image in $M_q, q \neq p$. Applying this algorithm to preimages in $H$ of every element in $H_p$, we conclude that every $H_p$ is a subgroup of $H$, and the conclusion follows. 
So our problem reduces to the problem of classifying the subgroups of $M_p$. First we make the following observation about subgroups of $\mathbb{Z}(p^{\infty})$. Every element of $\mathbb{Z}(p^{\infty})$ has the form $\frac{a}{p^n}$ where $(a, p) = 1$ and $n$ is unique; call $n$ the valuation $\nu_p \left( \frac{a}{p^n} \right)$. Every element of valuation $n$ generates the subgroup
$$P_n = \left\{ \frac{a}{p^n} : a \in \mathbb{Z} \right\} \cong \mathbb{Z}/p^n\mathbb{Z}$$
of $\mathbb{Z}(p^{\infty})$, and these subgroups are totally ordered by inclusion. It follows that in fact they are the only subgroups of $\mathbb{Z}(p^{\infty})$. 
Edit: Hum! So according to the comments this question is hopeless in full generality. Well, at least the constructions so far provide a large class of examples (obtained by taking direct sums of subgroups of $\mathbb{Z}(p^{\infty})$). 
