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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?

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    Below I addressed the general question. The case of Lie subgroups of $\mathbb{R}^n$ is straightforward; a simple classification based on dimension shows that only the discrete subgroups and the subspaces qualify. I believe any locally compact subgroup is necessarily closed, hence a Lie subgroup, so the same is true there.2012-02-09
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    Are you talking about vector spaces over $\mathbb R$ (or $\mathbb C$)? Otherwise some of those statements don't hold.2012-02-09
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    Sorry for forgetting to mention it: I mean a vector space over $\mathbb R$.2012-02-09
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    It's stated in [this MO thread](http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals) that already the classification of additive subgroups of $\mathbb R$ is "hopelessly complicated". Obviously this is a subproblem of your problem.2012-02-09
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    Many thanks to joriki and Qiaochu Yuan. This was quiet helpful. I must say I'm not that disappointed about the difficulty for a general classification (this was more a curiosity to me). But I would like to have a proof that all the possible Lie subgroups are like $\mathbb Z^k$ or like linear subspaces. I don't understand the dimension argument of Qiachou Yuan. And I would also like to see a proof that the Lie subgroups are exactly the locally compact subgroups. At least in this "simple" case somebody should have proven somewhere Hilbert's 5th problem :-).2012-02-10

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