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.
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.