I am trying to understand problem 8.5 in Kac's Infinite dimensional Lie algebras. It goes as follows.
Let $G$ be a semisimple algebraic group, let $\alpha$ be an automorphism of $G$ of order $m$, and let $\tilde{G}$ be the group of regular maps $\gamma: \mathbb{C}^{\times} \rightarrow G$. We have an action of $\alpha$ on $\tilde{G}$ by $\alpha(\gamma(t))=\gamma(\xi t)$ where $t$ is a primitive m'th root of unity. Show that for $g \in G$, the following are equivalent:
$g = \gamma^{-1} \alpha(\gamma)$ if and only if
$g \alpha(g) \alpha^2(g) \dots \alpha^{m-1}(g)=1$.
The forward direction is trivial, but I would appreciate it if someone would give me a hint on the reverse direction. In particular, given the second statement I have tried to produce maps $\gamma$ in two ways: either $\gamma$ maps $\mathbb{C}^{\times}$ to some point, or $\gamma(t)=(t \cdot Id) g$ or some variation of the above, where $t$ is a coordinate on $\mathbb{C}^{\times}$. I have not had success. Any hint would be appreciated
The second question is slightly deeper: I believe this is essentially showing that some Galois cohomology is 1: the formula $g \alpha(g) \alpha^2(g) \dots \alpha^{m-1}(g)=1$ looks pretty similar to the "trace" of $\alpha$, and I would guess that if I were looking for 1-cycles, I would be looking for something like this. Can someone either provide a nice reference or some orientation on how to understand the cohomology that is appearing here?
Thank you for any comments.