Let $L/K$ be a Galois extension with Galois group $G$. Suppose there exists $\alpha\in L$ and $\sigma\in G$ such that $L=K(\alpha)$ and $\sigma \alpha=\alpha^{-1}$. Prove that $[L:K]$ is even and $[K(\alpha+\alpha^{-1}):K]=\frac{1}{2}[L:K]$.
I think that I've almost figured this one out, but I can't seem to get one last detail. Here are my thoughts. Let $n$ be the order of $G$ and note that since $\sigma$ has order $2$ in $G$ we have $2\mid n=[L:K]$. In fact the Galois correspondence shows that $[L:L^{\langle \sigma \rangle}]=2$, so $[L^{\langle \sigma \rangle}:K]=n/2.$ My claim is now that $L^{\langle \sigma \rangle}=K(\alpha+\alpha^{-1})$. Clearly $\alpha+\alpha^{-1}\in L^{\langle \sigma \rangle}$ so this establishes $K(\alpha+\alpha^{-1}) \subseteq L^{\langle \sigma \rangle}$. Let's now suppose that this inclusion is strict, i.e., $K(\alpha+\alpha^{-1}) \subsetneq L^{\langle \sigma \rangle}$. We know that $K(\alpha+\alpha^{-1})=L^H$ for some $H\leq G$, and the strict inclusion says that there is some $\tau \in H\backslash \langle \sigma \rangle$ that fixes $\alpha+\alpha^{-1}$. I feel like I should be able to derive a contradiction from this (something along the lines of $\tau\alpha=\alpha$, which would imply $\tau=\sigma$), but I can't figure out where to go from here...
Since $\alpha$ satisfies the quadratic $T^2-T(\alpha + \alpha^{-1}) + 1$, we have that $[K(\alpha) : K(\alpha + \alpha^{-1})] \le 2$. Thus, $[K(\alpha + \alpha^{-1}) : K] \ge n/2$.
The tower $L^{\sigma}/K(\alpha + \alpha^{-1}) /K$ with $[L^{\sigma}:K] = n/2$ now implies $L^{\sigma} = K(\alpha + \alpha^{-1})$.