My Attempt: Let $\gamma_0=\sqrt{2+\sqrt{2}}, \gamma_1=-\sqrt{2+\sqrt{2}}, \gamma_2=\sqrt{2-\sqrt{2}}, \gamma_3=-\sqrt{2-\sqrt{2}}$. Let $L=\mathbb{Q}(\gamma_0, \gamma_1, \gamma_2, \gamma_3)$ and $M=\mathbb{Q}(\gamma_0)$. Then clearly $L$ is the splitting field of $f(t)=t^4-4t^2+2$ over $\mathbb{Q}$, hence $L: \mathbb{Q}$ is normal. Clearly $L: \mathbb{Q}$ is also finite, hence if we can show that $M^*$-the group of all $M$-automorphisms of L-is a normal subgroup of the Galois Group $G$ of $L$, then we can conclude that $M: \mathbb{Q}$ is normal by the Fundamental Theorem of Galois Theory.
Suppose $\tau\in G$. We wish to show that $\tau M^* \tau^{-1}\subset M^*$.
Proof:
Suppose $\sigma\in M^*$. We want to show that $\tau \sigma \tau^{-1}$ is a $M$-automorphism of L.
Clearly $\tau \sigma \tau^{-1}$ is an automorphism of L.
Suppose $x\in M$, then $x=\frac{p+q\gamma_0+r\sqrt{2}}{a+b\gamma_0+c\sqrt{2}}$, where $p, q, r, a, b, c \in \mathbb{Q}$. We want to show that $\tau \sigma \tau^{-1}(x)=x$. Dead end.
I then tried to apply the following Lemma:
"If $L: K$ is a field extension, $M$ is an intermediate field, and $\tau$ is a $K$-automorphism of L, then $\tau(M)^*=\tau M^*\tau^{-1}$".
If I can show that $M\subset\tau(M)$, then we'd have $\tau(M)^*\subset M^*$, and then we'd have $\tau M^*\tau^{-1}=\tau(M)^*\subset M^*$ by the lemma. This approach failed as well.
Any hint would be greatly appreciated.