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Given a vector space $V$ over a field $F$, if $F$ has (nonzero) characteristic $n$, then for any $v\in V$,

$$ \begin{align*} v+\cdots+v \textrm{ (n times)} &= (1_F+\cdots+1_F \textrm{ (n times)})\cdot v \\ &= 0_F\cdot v \\ &= 0 \end{align*} $$

That is, the vector space appears to "inherit" the characteristic of its field.

This seems to allow us to rule out the possibility of constructing a vector space using certain sets. For instance, even though $\mathbb{Z}_5\times\mathbb{Z}_2$ is an abelian group, it can't be a vector space over any field, since $(0,1)+(0,1)=(0,0)$ but $(1,0)+(1,0)\neq(0,0)$.

Is this correct? If so, is there a name for this idea? Is it related to any ideas with deeper significance?

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  • $\begingroup$ Any finite-dimensional vector space over a finite field has a number of elements which is the power of a prime. Indeed, let the ground field $\mathbb K$ have $p^r$ elements (for some prime $p$). Then the vector space is isomorphic to $\mathbb K^d$ where $d$ is the dimension, which has $p^{rd}$ elements $\endgroup$
    – Giulio R
    Nov 11, 2022 at 13:37

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a finite abelian group $G$ carry a vector space structure over a finite field $\mathbb{F}$ iff the order of any nontrivial element has order $Char(\mathbb{F})$, in which case it is isomorphic to $\mathbb{F}^{n}$, in particular a necessary condition is that $\lvert G \rvert =Char(\mathbb{F})^{n}$,So any abelian group whose order is not a power of some prime cannot admit a vector space structure over any finite field.

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  • $\begingroup$ In fact, the products of prime-order cyclic groups are the only finite groups that can be vector spaces (with the group operation as vector addition). $\endgroup$
    – Arthur
    Nov 11, 2022 at 14:00

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