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Let $F$ be a field. Determine the possible finite groups $G$ that are isomorphic to a subgroup of $F^+,$ the additive group of $F$.

This problem is a bit confusing for me. $F$ is an arbitrary field, and $F^+$ is an abelian group (finite or infinite). How can I determine the subgroups of a group about which I only know that it's abelian? Does the fact that the set $F-\{0\}$ is a multiplicative group (and other field axioms) tell me anything about $F^+$?

user557
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    Do you know about the characteristic of a field? – Alex Wertheim Jul 27 '18 at 01:05
  • Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $\mathbb F_p$? – user557 Jul 27 '18 at 01:12
  • Yep, you got it! And any finite additive subgroup will be a vector subspace over $\mathbb{F}{p}$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $\mathbb{F}{p}$.) – Alex Wertheim Jul 27 '18 at 01:14
  • @AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $\mathbb F_p$. – user557 Jul 27 '18 at 01:33
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    If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $\Delta$ be a finite additive subgroup of $F$, i.e. a finite $\mathbb{Z}$-submodule. Since every element of $\Delta$ is $p$-torsion, the $\mathbb{Z}$-module structure on $\Delta$ descends to a $\mathbb{Z}/p\mathbb{Z}$-module structure on $\Delta$. – Alex Wertheim Jul 27 '18 at 01:38
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    (Here, we are using the fact that if $M$ is an $R$-module, and $I$ is an ideal of $R$ such that $IM = 0$, then $M$ has the structure of an $R/I$-module such that for any $r \in R$, $r \cdot m = \overline{r} \cdot m$, where $\overline{r}$ is the class of $r$ in $R/I$.) – Alex Wertheim Jul 27 '18 at 01:40

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Hint: Let $G$ be a finite subgroup of $F^+$ and let $g \in G$ have order $n$. Consider $ng=0$ as the equation $(n\cdot1_F)\cdot g=0$ in the field $F$.

lhf
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  • In the field, $ng$ is the sum of $n$ copies of $g\in F$. If you mean I need to involve characteristic here, I'm not sure how exactly. – user557 Jul 27 '18 at 01:23
  • Then we could multiply both sides by $g^{-1}$ on the right to get $n\cdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude. – user557 Jul 27 '18 at 01:32
  • @user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$? – lhf Jul 27 '18 at 01:47
  • Assuming $pg$ means $(p\cdot 1_F)\cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $\mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(\mathbb F_{p^r},+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it? – user557 Jul 27 '18 at 01:59