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Let $V$ be an infinite-dimensional vector space over $\mathbb{Q}$.

Consider a proper subgroup $W$ of $V, +$ with the following property: each vector line $L$ (which we see as a subgroup of $V, +$) has infinite intersection with $W$ $$\vert L \cap W \vert \not\in \mathbb{N}.$$

Can such groups $W$ be classified in some way ? (Do they exist ?)

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    $\begingroup$ A subgroup satisfies this property iff it contains a basis. So it essentially amounts to classifying subgroups of $\mathbf{Q}^{(X)}$ containing $\mathbf{Z}^{(X)}$. There's a huge number of such subgroups. Say for $X$ countable, every countable abelian group of infinite countable rank occurs this way, and this is a huge class. $\endgroup$
    – YCor
    Commented May 7, 2021 at 17:14
  • $\begingroup$ @YCor : I think I am missing something. Say the dimension is $\vert \mathbb{N} \vert$. Let $W$ be the subgroup $\oplus_{\mathbb{N}}\mathbf{Z}$, and consider the vector line $L = \{ qv \vert q \in \mathbb{Q} \}$, with $v$ the vector $(1/p^i)_{i \in \mathbb{N}}$, $p$ some prime. If I am not mistaken, $W \cap L$ is the zero vector ? $\endgroup$
    – THC
    Commented May 7, 2021 at 18:05
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    $\begingroup$ @THC $v$ does not belong to your space. $\endgroup$
    – Mikael de la Salle
    Commented May 7, 2021 at 19:35
  • $\begingroup$ Two questions that might help: (1) Do maximal linearly independent subsets of W span V? (2) If B is a basis for V, then does the subgroup generated by B satisfies the property that you've given for W? (This is actually just a rewording of YCor's comment) $\endgroup$
    – Onur Oktay
    Commented May 8, 2021 at 8:21
  • $\begingroup$ @MikaeldelaSalle : you are right of course (I guess I was a bit sleepy). $\endgroup$
    – THC
    Commented May 8, 2021 at 23:51

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