Let $G$ be a finite subgroup of the invertible elements of a field $F$. Show that if char$F\neq0$, then $G$ is cyclic of order $n$ with $n$ prime to char$F$.

I have solved the first part but have no clue how to solve the coprime part.

Btw, this is a proposition in a step of a proof, so if you think there is any missing condition please let me know.

  • $\begingroup$ Too bad @Arthur deleted his answer. With your comment about $G$ being contained in a finite field, I think the problem's nailed. $\endgroup$ – Gerry Myerson Jun 26 at 9:26
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    $\begingroup$ Well I was thinking adjoining all the elements of $G$ to $F_p$ should yield a finite field of the form $F_{p^k}$. Not sure whether this is right though (I've forgotten a lot about abstract algebra) $\endgroup$ – trisct Jun 26 at 9:28
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    $\begingroup$ If $a$ is algebraic over a finite field $K$ then $K[a]$ is again finite. Since $G$ is finite then its every element is algebraic over the simple field $F_p$ and thus over any subfield of $F$. Therefore $F[G]$ is a finite subfield of $F$ containing $G$. Because we can chain finitely many finite extensions $F_p\subseteq F_p[g_1]\subseteq F_p[g_1][g_2]\subseteq\cdots\subseteq F[g_1]\cdots[g_m]$. $\endgroup$ – freakish Jun 26 at 9:59

Edit: with some of the comments above helping out by pointing me in the right direction, we can now complete the proof.

Let $p$ be the characteristic of $F$ and let $F_p$ be the minimal subfield of $F$. Then all elements of $G$ are algebraic over $F_p$, as they are all roots of the polynomial $x^{|G|} - 1$, which means that the chain of extensions $$ F_p\subseteq F_p(g_1)\subseteq F_p(g_1, g_2)\subseteq \cdots \subseteq F_p(g_1, g_2, \ldots, g_n) $$ are all algebraic extensions. The last field in the chain is therefore a finite subfield of $F$ that contains all the elements of $G$. Since $G$ is contained in some finite subfield of $F$, WLOG we may assume $F$ is finite.

$F$ has $p^k$ elements for some natural number $k\geq 1$, so the order of $G$ divides $p^k-1$ by Lagrange's theorem. By the Euclidean algorithm, $\gcd(p^k-1, p) = 1$. This shows that the order of $G$ and the characteristic of $F$ are coprime.

  • $\begingroup$ I don't think the field is supposed to be finite. Only the subgroup $G$, as I understand it. $\endgroup$ – Bernard Jun 26 at 9:13
  • $\begingroup$ What if $F$ is infinite? Can I conclude that $G$ must be contained in some finite extension of $F_p$ from $G$ being finite? $\endgroup$ – trisct Jun 26 at 9:14
  • $\begingroup$ A bit of nitpicking: I'd call $F_p$ just "(finite) prime field" and avoid further explanation. Also "they are all roots of the polynomial $x^{|G|}-1$", you didn't specify what "$n$" is. $\endgroup$ – freakish Jun 26 at 10:24
  • $\begingroup$ @freakish I want $F_p$ specifically to be a subfield of $F$, and I don't know if the standard $\Bbb F_p$ is that. It's isomorphic to one, sure, but it doesn't have to be one. And you're right about the $n$. I read the $n$ in the question above, and just continued using it without thinking too much about it. $\endgroup$ – Arthur Jun 26 at 10:31
  • $\begingroup$ @Arthur Oh, my bad, I've missed that $n$ was already defined. The notation $F_p$ is indeed misleading but you don't have to follow it. You can just define $F_p$ as the prime (as in minimal) subfield of $F$ and be done. It is then a simple lemma that $F_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. But that's really offtopic. Your answer is excelent! $\endgroup$ – freakish Jun 26 at 10:35

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