Finite Multiplicative Subgroup of Field is Cyclic
Theorem
Let $\left({F, +, \cdot}\right)$ be a field.
Let $C$ be a finite subgroup of $(F^*,\cdot)$.
Then $C$ is cyclic.
Proof
Write $|C| = p_1^{e_1} \cdots p_r^{e_r}$ with $p_i$, $i = 1,\ldots,r$ distinct primes (using the Fundamental Theorem of Arithmetic).
By the Fundamental Theorem of Finite Abelian Groups $C$ is a direct sum of subgroups $H_1,\ldots,H_r$ of orders $p_1^{e_1},\ldots, p_r^{e_r}$ respectively.
Let $p_i$ be a prime dividing $|C|$.
Then by [[$p$-group is cyclic]] $H_i$ has a generator $c_i$ of order $e_i$.
A few things here: 1. A $p$-group as defined in this site is a group whose order is the power of a prime. Then [[$p$-group is cyclic]] is disproved by the Klein 4-group (Viergruppe). 2. So assume that Group of Prime Order Cyclic meant instead. Then it needs to be shown how "$H_i$ has a generator $c_i$ of order $e_i$" follows from it.
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Finally by Group Direct Product of Cyclic Groups, since $\operatorname{gcd}(p_i^{e_i},p_j^{e_j}) = 1$ when $i \neq j$ we have that $C = H_1 \oplus \cdots \oplus H_r$ is cyclic.
The above would follow if all $H_i$ were actually cyclic, but I'm not sure they are: when $\vert H_i \vert$ exquals $4$ this is not necessarily the case, unless I've missed something obvious. In which case what's going on here needs to be made clear.
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$\blacksquare$
