## Every finite multiplicative subgroup of a field is cyclic

Let $F$ be a field and $G \leq F^\times$ be a finite multiplicative subgroup. Prove that $G$ is cyclic.

We prove this first in the case that $G$ is a $p$-group, say $|G| = p^k$, by induction on $k$. Given $G$, we let $\psi(d)$ denote the number of elements of $G$ of order $d$.

Suppose $|G| = p$; then certainly $G \cong Z_p$ is cyclic and $\psi(d) = \varphi(d)$ for each $d$ dividing $p$.

Suppose that every subgroup of order $p^t$ is cyclic for some $k \geq t \geq 1$ and that for all such $t$, $\psi(p^t) = \varphi(p^t)$. Let $G$ be a subgroup of order $p^{k+1}$. Note that the elements of $G$ are precisely the roots of $x^{p^{k+1}} - 1$ in $F$. Similarly, the roots of $x^{p^k} - 1$ are precisely those elements of $G$ whose order divides $p^k$. Suppose $\alpha$ and $\beta$ are two such roots; that is, $\alpha^{p^k} = 1$ and $\beta^{p^k} = 1$. Now $(\beta^{-1})^{p^k} = 1$, so that $(\alpha\beta^{-1})^{p^k} = 1$. Since $1^{p^k} = 1$, the set of roots of $x^{p^k} - 1$ (i.e. the set of elements of $G$ whose order divides $p^k$) form a subgroup of $F^\times$ of order $p^k$. By the induction hypothesis, this group is cyclic, and $\psi(p^t) = \varphi(p^t)$ for each $1 \leq t \leq k$. Note also that for each such $t$, $\psi(p^t)$ with respect to $G$ is equal to $\psi(p^t)$ with respect to this subgroup. Since $\sum_{t=1}^{k+1} \psi(p^t) = p^t = \sum_{t=1}^{k+1} \varphi(p^t)$, we have $\psi(p^{k+1}) = \varphi(p^{k+1})$. In particular, $G$ contains an element of order $p^{k+1}$, and so is cyclic.

Thus every finite $p$-subgroup of $F^\times$ is cyclic. Now let $G$ be an arbitrary finite subgroup; since $G$ is abelian, it is the direct product of its $p$-subgroups. Since these are cyclic, $G$ is cyclic.