Fixed field of $\mathbb{Q}(\xi)$ over $\big\lbrace \operatorname{Id}, \sigma^3, \sigma^6, \sigma^9 \big\rbrace$ What is the fixed field of $\mathbb{Q}(\xi)$ (as extension over $\mathbb{Q}$) over $H := \big\lbrace \operatorname{Id}, \sigma^3, \sigma^6, \sigma^9 \big\rbrace \leq S_3$ where $\sigma : \xi \mapsto \xi^2$ and $\xi = e^\frac{2\pi i}{13}$?

What I've done so far:
I first determined the permutation table for $\xi$ over $\sigma$:





$\operatorname{Id}$
$\sigma$
$\sigma^2$
$\sigma^3$
$\sigma^4$
$\sigma^5$
$\sigma^6$
$\sigma^7$
$\sigma^8$
$\sigma^9$
$\sigma^{10}$
$\sigma^{11}$




$\xi$
$\xi$
$\xi^2$
$\xi^4$
$\xi^8$
$\xi^3$
$\xi^6$
$\xi^{12}$
$\xi^{11}$
$\xi^9$
$\xi^5$
$\xi^{10}$
$\xi^{7}$




Then I calculated the fixed field for every element of $H$ (by comparing linear combination scalars):




$\operatorname{Id}$
$\sigma^3$
$\sigma^6$
$\sigma^9$




$\mathbb{Q}(\xi)$
$\mathbb{Q}(\xi^2 + \xi^3 + \xi^{10} + \xi^{11}, \xi^4 + \xi^6 + \xi^7 + \xi^9)$
$\mathbb{Q}(\xi^2 + \xi^{11}, \xi^3 + \xi^{10}, \xi^4 + \xi^9, \xi^5 + \xi^8 + \xi^6 + \xi^7)$
$\mathbb{Q}(\xi^2 + \xi^{10} + \xi^{11} + \xi^3, \xi^4 + \xi^7 + \xi^9 + \xi^6)$




I'm not sure how to continue from here; is the entire fixed field just the intersection of these smaller ones?

If my above hypothesis is true, is it also true in general? i.e. is
$$\mathbb{L}^H = \bigcap_{\sigma \in H} \mathbb{L}^\sigma$$
for any Galois extension $\mathbb{L}/\mathbb{K}$ and a subgroup of the Galois group $H \leq \operatorname{Gal}(\mathbb{L}/\mathbb{K})$ true?
My intuition tells me it is true. However, I do not know how I would prove it formally. Would I use the definition of a fixed field $\mathbb{L}^H = \big\lbrace x \in \mathbb{L} \mid \sigma(x) = x \quad \forall \sigma \in H \big\rbrace$ and argue that if it's in the intersection then it's fiexd by all elements of $H$?
I've now moved this hypothesis into a separate question
 A: For any integer $k\in[0,12],$ $\sigma^3(\xi^k)=\xi^{8k\bmod{13}}=\xi^{\sigma(k)}$ where $\sigma=(1\;8\;12\;5)(2\;3\;11\;10)(4\;6\;9\;7)$ hence the subfield fixed by $\sigma^3$ is
$$K=\Bbb Q(\alpha,\beta,\gamma)$$
where
$$\alpha=\xi+\xi^8+\xi^{12}+\xi^5,\;\beta=\xi^2+\xi^3+\xi^{11}+\xi^{10},\;\gamma=\xi^4+\xi^6+\xi^9+\xi^7.$$
But $1+\alpha+\beta+\gamma=\sum_{k=0}^{12}\xi^k=0$ hence e.g. $\gamma$ can be eliminated, so $K=\Bbb Q(\alpha,\beta).$ And $\beta=-\alpha^2-2\alpha+2$ can also be eliminated. Finally,
$$K=\Bbb Q(\alpha).$$
The subfield fixed by $H$ is the same, since every number fixed by $\sigma^3$ is fixed by any power of it.
Edit:

*

*The relation $\beta=-\alpha^2-2\alpha+2$ was found by squaring $\alpha$ and grouping the various powers of $\xi$:$$\alpha^2=\beta+2\gamma+4=\beta+2(-1-\alpha-\beta)+4=-2\alpha+2-\beta.$$

*The same method allows to compute the minimal polynomial of $\alpha$:$$\alpha^3=\alpha(-2\alpha+2-\beta)=-2\alpha^2+2\alpha-(\alpha+2\beta+\gamma)$$$$=-2\alpha^2+2\alpha-\beta+1=-\alpha^2+4\alpha-1.$$
