$K^\times$ isomorphic to $\mathbb{Z}/2n\mathbb{Z}$ when $K^\times$ is cyclic Let $K$ be a field so that $K^\times$ is cyclic. Assume $\operatorname{char} K \neq 2$. Prove that $K$ is finite and $K^\times$ is isomorphic to $\mathbb{Z}/2n\mathbb{Z}$ for some $n$.
To prove that $K$ is finite, I said that if it's not finite, it must contain $\mathbb{Q}$ and so does $K^\times$, threfore it cannot be cyclic. Does that make any sense?
About the second part, I'd be happy for a hint.
Thanks.
 A: To prove that $K$ is finite, we assume by way of contradiction that $K^{\times} = \langle a \rangle$ is infinite cyclic. If $(a^{i})^{2} = 1$, then $2 i = 0$ so that $i=0$. It follows that $-1 = 1$ in $K$, and the characteristic is $2$. This would be excluded by assumption, but we can prove this does not happen even in characteristic $2$. If $a$ is algebraic over $F = \mathbb{Z}/2\mathbb{Z}$, then we get that $K^{\times}$ contains a finite cyclic subgroup $F(a)^{\times}$, a contradiction. Thus $a$ is trascendental over $F$, but then there's no way we can obtain $1+a$ as a power of $a$.
A: Once you've ruled out infinite fields, then you want to turn your focus towards proving: If $K$ is finite such that $char(K) > 2$, then $K^\times \cong \mathbb{Z}_{2n}$ for some $n \in \mathbb{N}$.  Here's the simplest way I know how to approach this:
If $K$ is a finite field isomorphic to $\mathbb{F}_{p^m}$, then $K^\times = K$ \ $\{0\}$, which has $p^m - 1$ elements.  Note that, since $char(K) > 2$, $p$ is an odd prime, and so $p^m - 1$ is even.  Therefore, it will suffice to show that $K^\times \cong \mathbb{Z}_{p^m -1}$.
First, consider an abelian group $G$ and two elements $a, b \in G$.  Show that $\gcd(|a|, |b|) = 1 \implies |ab| = |a| \cdot |b|$.  
Next, consider $m = \max \{|g| : g \in G \}$.  Show that, for any arbitrary $x \in G$, $|x|$ divides $m$.  
Once you have this result, then note that $K^\times$ is an abelian group under multiplication.  If we let $m$ be defined as above, consider the polynomial $f(x) = x^m - 1$.  From our work above, we know that every element of $K^\times$ is a root of this polynomial.  So we'll have $\deg(f) = m \geq |K^\times|$ by the fundamental theorem of algebra.  However, by our definition of $m$ being the order of an element in $K^\times$, then by Lagrange, we conclude...
We therefore have a generator for $K^\times$, so it is cyclic.
A: Your idea with $\mathbb Q^\times$ is in the right direction: Let $F\subseteq K$ be the prime field (so $F\cong \mathbb Q$ or $F\cong \mathbb F_p$ for some prime $p$). Then $F^\times$ is a subgroup of the cyclic group $K^\times$, hence cyclic. We know that $\mathbb Q^\times$ is not cyclic (if $\mathbb Q^\times=\langle a\rangle$, then we must have $a\in\{2,\frac12\}$ because all square or higher roots of $2$ are irrational; by the same argument, $a\in\{3,\frac13\}$, contradiction), hence conclude that the characteristic is nonzero.
Also, the only finite subgroup of an infinite cyclic group is the trivial one, but $\mathbb F_p^\times$ is a group of order $p-1\ne1$ because $p\ne 2$.
We conclude that $K$ is finite. Then $K$ is isomorphic to some $\mathbb F_{p^m}$ and $|K|=p^m-1$ for some $m\ge1$ and odd prime $p$. So finally $K^\times\cong\mathbb Z/2n\mathbb Z$ with $n=\frac{p^m-1}{2}\in\mathbb N$.
