Let $\zeta \in \mathbb{C}$ a primitive $p^{th}$ root of unit ($\zeta^p=1$ holds and no smaller power works) with $p$ an odd primeand assume $p > 2$. Consider $E = \mathbb{Q}(\zeta)$. This cyclotomic extension is a Galois extension with $ G = \mathrm{Gal}(E/\mathbb{Q}) = \left( \mathbb{Z} / p \mathbb{Z} \right)^{\times}$ so $G$ is the multiplicative group $ G = \left( \{ 1 , 2 , ... , p-1 \} , \cdot \bmod p \right)$. Let $H$ be a subgroup of $G$ of order 2. Define $ \alpha = \sum_{i \in H} \zeta^i \quad \mbox{ and } \quad \beta = \sum_{i \in G-H} \zeta^i \ .$ I showed that $\alpha$ and $\beta$ are fixed under $H$ and that $\alpha$ and $\beta$ are the roots of $x^2 + x + \alpha \beta \in \mathbb{Q}[x]$.
I want to calculate $\alpha \beta$ and from that deduce that the fixed field of $H$ (that is, $E^H$) is $E^H = \mathbb{Q}(\sqrt{p})$ when $p = 1 \bmod 4$ and $E^H = \mathbb{Q}(\sqrt{-p})$ when $p = 3 \bmod 4$.
We have that $H = \{ \pm 1 \}$ since $(p-1)^2 \stackrel{p}{\equiv} (-1)^2 = 1$ and thus it is an element of order $2$. I have calculated $\alpha = \zeta + \zeta^{p-1} = \zeta + \zeta^{-1}$ and $\beta = \zeta^2 + ... + \zeta^{p-2}$ by multiplying them and thus use the formula for geometric series but did not reach a simple expression, which relates to the second part of the question.
I would appriciate help.