# Order of a finite multiplicative subgroup of a field

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.

• Too bad @Arthur deleted his answer. With your comment about $G$ being contained in a finite field, I think the problem's nailed. – Gerry Myerson Jun 26 at 9:26
• 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) – trisct Jun 26 at 9:28
• 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 $F_p\subseteq F_p[g_1]\subseteq F_p[g_1][g_2]\subseteq\cdots\subseteq F[g_1]\cdots[g_m]$. – freakish Jun 26 at 9:59

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.
• I don't think the field is supposed to be finite. Only the subgroup $G$, as I understand it. – Bernard Jun 26 at 9:13
• What if $F$ is infinite? Can I conclude that $G$ must be contained in some finite extension of $F_p$ from $G$ being finite? – trisct Jun 26 at 9:14
• 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. – freakish Jun 26 at 10:24
• @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. – Arthur Jun 26 at 10:31
• @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! – freakish Jun 26 at 10:35