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Let $G$ be a finite abelian group and let $V$ be a finite-dimensional irreducible representation of $G$ over a field $k$ of characteristic $0$. Is it the case that the action of $G$ on $V$ factors through a cyclic group?

This is easy if $k$ is algebraically closed, and I'm pretty sure it is true in general (but I can't seem to prove it or find a suitable reference).

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    $\begingroup$ You need the result that every finite subgroup of the multiplicative group of a field is cyclic. The easiest way to prove this is by using that a polynomial of degree $d$ has at most $d$ roots and that a finite abelian group is cyclic if and only if it's order is its exponent. $\endgroup$
    – Simon Wadsley
    Commented Jan 17 at 19:14
  • $\begingroup$ (To apply the previous comment you need to observe that the subalgebra generated by the image of $G$ is a field.) $\endgroup$
    – YCor
    Commented Jan 17 at 19:16
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    $\begingroup$ Ah, I see: the point is that since $G$ is abelian, the image of $G$ in $End(V)$ lies in $End_G(V)$, which is a division ring. You then argue that the subalgebra generated by the image of $G$ in $End_G(V)$ is a commutative division ring (I guess you have to show that it is closed under inverses?), and thus a field. It then follows from the standard fact that Simon quotes. Is this a correct summary? $\endgroup$
    – Miranda
    Commented Jan 17 at 19:22
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    $\begingroup$ @Miranda for nonzero $f$ in this subalgebra, $f^{-1}$ is a polynomial in $f$ so is in the subalgebra too. $\endgroup$
    – YCor
    Commented Jan 17 at 21:08
  • $\begingroup$ @YCor: Ah, good point, thanks to all! $\endgroup$
    – Miranda
    Commented Jan 17 at 21:39

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