Vtyjkyuk

Let $L/K$ be a Galois extension with Galois group $G$. Suppose there exists $alphain L$ and $sigmain G$ such that $L=K(alpha)$ and $sigma alpha=alpha^-1$. Prove that $[L:K]$ is even and $[K(alpha+alpha^-1):K]=frac12[L:K]$.

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 $2mid n=[L:K]$. 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 $alpha+alpha^-1in L^langle sigma rangle$ 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 $Hleq G$, and the strict inclusion says that there is some $tau in Hbackslash langle sigma rangle$ that fixes $alpha+alpha^-1$. I feel like I should be able to derive a contradiction from this (something along the lines of $taualpha=alpha$, 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 $L^sigma = K(alpha + alpha^-1)$.