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Does there exist a finite abelian group $G$, such that for no field $F$ is it the case that $G$ is isomorphic to either the additive group of $F$ nor the multiplicative group of non-zero elements of $F$? I would prefer a counterexample of minimum cardinality.

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  • $\begingroup$ The cyclic group of order $9$ is not one. $\endgroup$ Commented Jul 4, 2023 at 23:32

3 Answers 3

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The two results being used below are that the multiplicative group of a finite field is cyclic and the classification of finite abelian groups. For the former, see this MO thread for a discussion of some accessible proofs.


I made a really bad mistake in the first version of this answer! Since this answer's been accepted I can't delete it, but the fix below is due to Fishbane's comment so I've made this CW.

$C_2\times C_4$ is not the additive or multiplicative group of any field. Meanwhile, it's easy to check that every abelian group of size $<8$ is either the additive or multiplicative group of a field:

  • For each prime $p$ (here $p=2,3,5,7$), $C_p$ is the additive group of the field $\mathbb{F}_p$.

  • The only other abelian groups are the trivial group (= multiplicative group of $\mathbb{F}_2$), the Klein four group (= additive group of $\mathbb{F}_4$), and $C_6$ (= multiplicative group of $\mathbb{F}_7$).

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  • $\begingroup$ You probably want $n=14$, $15+1=16=2^4$. $n=8$ also doesn't work. $8+1=9=3^2$ and $8=2^3$ $\endgroup$
    – Fishbane
    Commented Jul 4, 2023 at 23:39
  • $\begingroup$ @Fishbane Yup, finger slip. Fixed! $\endgroup$ Commented Jul 4, 2023 at 23:39
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    $\begingroup$ $C_2\times C_2\times C_2$ is the additive group of the field of order $8$ also $C_8$ is the multiplicative group of the field of order $9$. The smallest $n$ where all groups of order $n$ are not the additive or multiplicative group of a field is $n=14$. $C_2 \times C_4$ is indeed a valid answer to the question though. $\endgroup$
    – Fishbane
    Commented Jul 4, 2023 at 23:48
  • $\begingroup$ @Fishbane Gah, this isn't my day ... $\endgroup$ Commented Jul 5, 2023 at 0:10
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    $\begingroup$ @DanielR.Collins Gaaaaahhhhh! $\endgroup$ Commented Jul 5, 2023 at 15:24
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The group $\mathbb{Z_2}\times\mathbb{Z_6}$ for example. It can't be the multiplicative group of any (necessary finite) field because it is not cyclic. And it can't be the additive group of any field because its order is not a power of a prime.

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There is a slightly easier argument, which does not require knowledge of the groups of finite fields. Any group of order $14$ cannot be the additive or multiplicative group of a field since neither $14$ nor $15$ is a prime power.

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    $\begingroup$ Just to make explicit, the one fact needed here is that any finite field has size a prime power, $p^q$; so its additive group has size $p^q$, and its multiplicative group has size $p^q-1$. $\endgroup$ Commented Jul 5, 2023 at 18:13

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