Intermediate fields of splitting field I'm trying to find the intermediate fields of the extension $\mathbb Q\big /\mathbb Q(\alpha)$, where $\alpha = \sqrt{7+\sqrt{13}}$. To do so I've tried to use the Galois correspondence. I've already found that $\rm{Gal}\left(\mathbb Q\big /\mathbb Q(\alpha)\right)$ has order $4$ and that is isomorphic to $\mathbb Z\big/2\mathbb Z\times \mathbb Z\big/2\mathbb Z$, which has three subgroups of order 2. Therefore, there must be three normal intermediate fields in the extension.
$E:=\mathbb Q(\alpha)$ is the splitting field of $f(x)=x^4-14x^2+36$, which has $4$ roots numbered as:
$$\left\{\alpha_1 = \alpha, \alpha_2 = -\sqrt{7+\sqrt{13}}, \alpha_3 = \sqrt{7-\sqrt{13}}, \alpha_4 = -\sqrt{7-\sqrt{13}}\right\}.$$
Then, the automorphisms in $\rm{Gal}\left(\mathbb Q\big /\mathbb Q(\alpha)\right)$ are
$$\sigma_1(\alpha) = \alpha_1$$
$$\sigma_2(\alpha) = \alpha_2$$
$$\sigma_3(\alpha) = \alpha_3$$
$$\sigma_4(\alpha) = \alpha_4$$
EDIT: corrected subgroups
If we see them as elements of $S_4$, they are $id, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)$, respectively. Thus, the subgroups are $H_1:=\langle(1,2)(3,4)\rangle, H_2:=\langle(1,3)(2,4)\rangle$ and $H_3:=\langle(1,4)(2,3)\rangle$ (all isomorphic to $\mathbb Z\big/2\mathbb Z$). 
Then, to find the intermediate fields I'm looking for $E^{H_i}=\{x\in E\mid \sigma(x), \forall\sigma\in H_i\}$
However, when I try with $H_2$, for example, it gets very nasty. In this case we'd have to impose that $\sigma_2(\gamma)=\gamma$ for all $\gamma\in E$. On the one hand, since $\{1,\alpha,\alpha^2,\alpha^3\}$ is a $\mathbb Q-$basis for $E$, 
$$\gamma = a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3,\qquad a_i\in\mathbb Q$$
On the other hand, we have
$$\sigma_2(\gamma) = a_0+a_1\sigma_2(\alpha)+a_2\sigma_2(\alpha^2)+a_3\sigma_3(\alpha^3)$$
$$=a_0-a_1\alpha+a_2\alpha^2-a_3\alpha^3$$
And then, coefficients of both expressions should be equal.
EDIT: added manipulations of these expressions
From these expressions we get that $\alpha_1=\alpha_3=0$ and $\alpha_0,\alpha_2$ are free (so it has 2 degrees of freedom as expected). 
$$\Rightarrow E^{H_2}=\{a+b(7+\sqrt{13})\mid a, b\in \mathbb Q\}=\mathbb Q(7+\sqrt{13})=\mathbb Q(\sqrt{13}).$$
So, $\mathbb Q(\sqrt{13})$ is the intermediate field that corresponds to the group $H_2$.
But I don't know how to do it with the other subgroups..
With $\sigma_3(\alpha) = \alpha_3 = \frac{6}{\alpha}$,
$$\sigma_3(\gamma) = a_0+a_1\frac{6}{\alpha}+a_2\left(\frac{6}{\alpha}\right)^2+a_3\left(\frac{6}{\alpha}\right)^3$$
 A: There is a general method to deal with these sorts of extensions. Suppose we have $F$ a field of characteristic not equal to $2$ and suppose we adjoin $\alpha = \sqrt{a + b\sqrt{c}}$ Suppose that $b\neq0$, $a^2 - b^2c = h^2$ for some $h\in F$. Let $f(x)$ be the minimal polynomial of $\sqrt{a + b\sqrt{c}}$ (put some nice conditions on $a,b,c$ so that $F(\alpha)$ is an extension of degree $4$).
Let us call $\alpha_1 = \alpha$ , $\alpha_2 = \sqrt{a - b\sqrt{c}}$, $-\alpha_1= \alpha_3$ and $-\alpha_2 = \alpha_4$. We choose the following presentation for $V_4$ (the Klein 4- group isomorphic to $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$): $$V_4 = \{e, (12)(34), (13)(24),(14)(23)\}.$$ Note that $(14)(23) = \big((13)(24)\big)\big((12)(34)\big)$. Then we can see that each cycle in $V_4$ acts on the roots according to the numbering convention: For example the cycle $(12)(34)$ sends  $\alpha_1 \mapsto \alpha_2$, $\alpha_3\mapsto \alpha_4$. Now for the sake of simplicity let us call 
$$\theta = (12)(34), \hspace{5mm} \gamma = (13)(24), \hspace{5mm} \gamma\theta = (14)(23).$$
Then it is easy to see that $\theta(\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2, \gamma\theta(\alpha_1 - \alpha_2) = \alpha_1 - \alpha_2$. Now because $\gamma(\alpha_1) = \alpha_3$, squaring both sides we see that $\gamma(\sqrt{c} ) = \sqrt{c}$. It follows from these observations that
$$F(\alpha_1 + \alpha_2) \subset E^{\langle \theta \rangle}, \hspace{5mm} F(\sqrt{c}) \subset E^{\langle \gamma \rangle}, \hspace{5mm} F(\alpha_1 - \alpha_2) \subset E^{\langle \gamma\theta\rangle }.$$
By computing minimal polynomials/degrees we get that these subset inclusions are actually equalities and so you are done.
Edit: As Jyrki points out, the only presentation for $V_4$ that we can choose is the one given in my answer as that presentation is the only one which gives a transitive subgroup of $S_4$.
A: Perhaps it is worth mentioning that $\mathbf{Q}(\alpha) = \mathbf{Q}\left(\sqrt{\dfrac{13}{2}}, \sqrt{\dfrac{1}{2}}\right)$. In fact, as in the post of @BenjaLim, if $a^2 - b^2 c$ is a square (in $\mathbf{Q}$, in this case), you can solve
\begin{equation}
\sqrt{a + b \sqrt{c}} = \sqrt{u} + \sqrt{v}
\end{equation}
for $u, v \in \mathbf{Q}$, as they are the solutions of $x^2 - a x + \dfrac{b² c}{4} = 0$. Rewriting the extension this way makes the computations of the Galois group and of the intermediate fields slightly easier.
Thus, we are regarding the extension  $\mathbf{Q}\left(\sqrt{\dfrac{13}{2}}, \sqrt{\dfrac{1}{2}}\right) / \mathbf{Q}$ as the splitting field of the reducible polynomial
\begin{equation}
\left(x^2 - \dfrac{13}{2}\right) \cdot \left(x^2 - \dfrac{1}{2}\right),
\end{equation}
whose roots are
\begin{equation}
\beta_{1} = \sqrt{\frac{13}{2}}, \quad
\beta_{2} = - \sqrt{\frac{13}{2}}, \quad
\beta_{3} = \sqrt{\frac{1}{2}}, \quad
\beta_{4} = - \sqrt{\frac{1}{2}}.
\end{equation}
If you regard the Galois group as a permutation group acting on the indices of the $\beta_i$, it takes the form $\{1,(12),(34),(12)(34)\}$. Not a transitive group, but the roots are those of a reducible polynomial here.
The intermediate subfields of course are
\begin{equation}
\mathbf{Q}\left(\sqrt{\dfrac{13}{2}}\right), \quad
\mathbf{Q}\left(\sqrt{\dfrac{1}{2}}\right), \quad
\mathbf{Q}\left(\sqrt{\dfrac{13}{2}} \cdot \sqrt{\dfrac{1}{2}}\right)
=
\mathbf{Q}\left(\sqrt{13}\right)
.
\end{equation}
EDIT In a previous version I had made an incorrect claim. Thanks to @Kits89 for letting me know in his comment below.
