Remember that two cocycles $a, a^\prime\colon G\to L^*$ are called cohomologous, denoted by $a\sim a^\prime$ , if there exists $b\in L^*$ , such that $a^\prime(\tau)=ba(\tau)\tau(b^{-1})$ for all $\tau\in G$ . Then $$H^1(G,L^*)=\{a\colon G\to L^*|a\text{ is a cocycle}\}/\sim.$$ Now let $a\colon G\to L^*$ be a cocycle. Then consider the map $$\alpha\colon L\to L, c\mapsto\sum_{\tau\in G}a(\tau)\tau(c).$$ Since elements of the Galois group are linearly independent, $\alpha$ is not $0$ . So we can choose $c\in L$ , such that $b=\alpha(c)\neq 0$ . Then for $\sigma\in G$ we have

Now we prove the corollary. Denote the norm by $N$ . Now if $x=\frac{y}{\sigma(y)}$ , we have $$N(x)=N\left(\frac{y}{\sigma(y)}\right)=\prod_{\tau\in G}\frac{\tau(y)}{\tau(\sigma(y))}=1.$$ Now let $N(x)=1$ , $n=|G|$ . Since $G$ is assumed cyclic, let $\sigma$ be a generator of $G$ . $G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ . We define the map $\tilde{x}\colon \mathbb{Z}/n\mathbb{Z}\to L^*$ by $$\tilde{x}([i])=\prod_{0\leq j\leq i-1}\sigma^j(x),$$ where $[i]$ denotes the class of $i\in\mathbb{Z}$ in $\mathbb{Z}/n\mathbb{Z}$ . Since $N(x)=1$ , $\tilde{x}$ is well defined. We have