Lemma 1

Let G be a finite subgroup of $K^*$ of order $m$; then G is cyclic and we have $G = \mu_m(K)$.

Consider G as Z-module; we have $mx = 0$ for all $x \in G$, hence the annihilator of G is an ideal of the form $r\mathbf{Z}$, where the integer $r \geq 1$ divides $m$. We thus have $G \subset \mu_r(K)$. By Lemma 4 (V, p. 74) applied with $A = \mathbf{Z}$ and $M = G$ there exists an element $x$ of G of order $r$; let $G'$ be the cyclic subgroup of G generated by $x$. We have $G' \subset \mu_r(K)$, Card $(G') = r$ and Card $\mu_r(K) \leq r$; therefore we have

G' = μ_r(K) ⊃ G and G is cyclic of order r, hence equal to μ_r(K). Since G has order m, we have m = r and the lemma follows.