Galois group of $(X^4 - 2)(X^2 + 2)$ I'm trying to compute the Galois group of the polynomial $P(X) = (X^4 - 2)(X^2 + 2)$, but I think that I made a mistake. This is what I thought:

$P(X)$ has roots $\pm i \sqrt{2}, \pm i \sqrt[4]{2}, \pm \sqrt[4]{2}$, then the splitting field of $P(X)$ is $K := \mathbb{Q}[\sqrt[4]{2}, i \sqrt[4]{2}, i \sqrt{2}]$, but $\sqrt{2} = \left( \sqrt[4]{2} \right)^2 \in \mathbb{Q}[\sqrt[4]{2}, i]$, $i \sqrt{2} \in \mathbb{Q}[\sqrt[4]{2}, i]$, $i \sqrt[4]{2} \in \mathbb{Q}[\sqrt[4]{2}, i]$ and $i = (i \sqrt[4]{2})^3 (\sqrt[4]{2}) \left( \frac{-1}{2} \right) \in K$, then $K = \mathbb{Q}[\sqrt[4]{2}, i]$.
Observing that $X^4 - 2$ is monic, has $\sqrt[4]{2}$ as a root and is irreducible on $\mathbb{Q} [X]$ by Eisenstein's criterion applied for $p = 2$, we have that $X^4 - 2$ is a minimal polynomial over $\mathbb{Q}$ which has a root $\sqrt[4]{2}$, then $\left[ \mathbb{Q} [\sqrt[4]{2}] : \mathbb{Q} \right] = 4$. Furthermore,  $X^2 + 1$ is monic, which has a root $i$ and is irreducible over $\mathbb{Q}[\sqrt[4]{2}] [X]$ since $X^2 + 1$ is a polynomial of degree $2$ and $i \notin \mathbb{Q}[\sqrt[4]{2}]$, then $X^2 + 1$ is a minimal polynomial over $\mathbb{Q}[\sqrt[4]{2}] [X]$, then $\left[ \mathbb{Q} [\sqrt[4]{2},i] : \mathbb{Q} [\sqrt[4]{2}] \right] = 2$.
By the tower law,
$$\left[ \mathbb{Q} [\sqrt[4]{2},i] : \mathbb{Q} \right] = \left[ \mathbb{Q} [\sqrt[4]{2},i] : \mathbb{Q} [\sqrt[4]{2}] \right] \ \left[ \mathbb{Q} [\sqrt[4]{2}] : \mathbb{Q} \right] = 2 \cdot 4 = 8,$$
but there are only $4$ automorphisms of $\mathbb{Q} [\sqrt[4]{2},i]$ which fixes $\mathbb{Q}$ and this contradicts the fact that the cardinality of the Galois group of $K$ over $\mathbb{Q})$ is equal to $\left[ \mathbb{Q} [\sqrt[4]{2},i] : \mathbb{Q} \right]$

What am I doing wrong?
 A: You are wrong that there are only $4$ automorphisms of $\mathbb{K}:=\mathbb{Q}(\sqrt[4]{2},\text{i})$ that fix $\mathbb{Q}$.  The automorphism group $G:=\text{Aut}_\mathbb{Q}(\mathbb{K})$ is isomorphic to the unique nonabelian semidirect product $C_4\rtimes C_2\cong D_4$, where $C_k$ is the cyclic group of order $k$ and $D_k$ is the dihedral group of order $2k$. The group $G$ is generated by $\tau$ and $\sigma$ where $$\tau(\sqrt[4]{2})=\text{i}\sqrt[4]{2}\,,\,\,\tau(\text{i})=\text{i}\,,\,\,\sigma(\sqrt[4]{2})=\sqrt[4]{2}\,,\text{ and }\sigma(\text{i})=-\text{i}\,.$$
Note that $\langle \tau\rangle\cap\langle \sigma\rangle=\text{Id}$, with $\langle \tau\rangle\cong C_4$ and $\langle \sigma\rangle\cong C_2$.  In addition,
$$(\tau\circ\sigma\circ\tau)(\sqrt[4]{2})=(\tau\circ \sigma)(\text{i}\sqrt[4]{2})=\tau(-\text{i}\sqrt[4]{2})=\sqrt[4]{2}=\sigma(\sqrt[4]{2})$$
and 
$$(\tau\circ\sigma\circ\tau)(\text{i})=(\tau\circ \sigma)(\text{i})=\tau(-\text{i})=-\text{i}=\sigma(\text{i})\,.$$
Hence, $\tau\circ\sigma\circ\tau=\sigma$, or $\tau\circ\sigma=\sigma\circ\tau^{-1}$.  That is,
$$G=\big\langle \tau,\sigma\,\big|\,\tau^{\circ 4}=\text{id}\,,\,\,\sigma^{\circ 2}=\text{id}\,,\text{ and }\tau\circ\sigma=\sigma\circ\tau^{-1}\rangle\cong D_4\,.$$
