# 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?

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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. –  Qiaochu Yuan Feb 9 '12 at 18:46
Are you talking about vector spaces over $\mathbb R$ (or $\mathbb C$)? Otherwise some of those statements don't hold. –  joriki Feb 9 '12 at 20:07
Sorry for forgetting to mention it: I mean a vector space over $\mathbb R$. –  Hans Feb 9 '12 at 20:09
It's stated in this MO thread that already the classification of additive subgroups of $\mathbb R$ is "hopelessly complicated". Obviously this is a subproblem of your problem. –  joriki Feb 9 '12 at 20:16
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 :-). –  Hans Feb 10 '12 at 3:04

## 1 Answer

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 $\mathbb{Z}(p^{\infty})$).

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Haha, the solution to this problem appears to be a lot more complicated than the statement of the original question. –  Eric Haengel Feb 9 '12 at 18:54
"subgroups of a direct sum must be direct sums of subgroups" is probably not true in the way you want it to be true. Many subgroups of Q^n are indecomposable but generate vector spaces of dimension n, and so are definitely not direct sums of subgroups of Q. Subgroups of Q need not contain 1. –  Jack Schmidt Feb 9 '12 at 19:04
@Jack: oops. You're right, of course. –  Qiaochu Yuan Feb 9 '12 at 19:15