Let $K = \mathbb{Q}_2$ and $d_2 = 3x^4 + 12 x^2 + 4x - 4 \in K[x]$. Let $L_0$ be the splitting field of $d_2$ and let $\alpha \in L_0$ be a root of $d_2$. Let $L/L_0$ be the quadratic extension obtained by $\delta := (1+2\alpha + \alpha^3)^{1/2}$.
In this paper (more precisely in Example 4.2), it is claimed that the Galois group of $L/K$ is $\operatorname{GL}_2(\mathbb{F}_3)$ and I have trouble understanding why.
I know that by the theory of Galois groups of quartic polynomials, it is $\operatorname{Gal}(L_0/K) = S_4$. And obviously, it is $\operatorname{Gal}(L/L_0) = C_2$. But $\operatorname{GL}_2(\mathbb{F}_3)$ is quite a big group (has order $48$) and I have problems understanding where the generators of $L/K$ (i.e. the roots of $d_2$, and $\delta$) map. I know that $\delta \mapsto -\delta$ generates $\operatorname{Gal}(L/L_0) = C_2$ and $\phi_1,\phi_2 \in \operatorname{Gal}(L_0/K)$ with $\phi_1(\alpha_i) = \alpha_{i+1 \mod 4}$ and $\phi_2(\alpha_1) = \alpha_2$, $\phi(\alpha_2) = \alpha_1$, $\phi(\alpha_j) =\alpha_j$ for $j=3,4$ generate $\operatorname{Gal}(L_0/K)$ (here, $\alpha=\alpha_1, \dots, \alpha_4$ are the four distinct roots of $d_2$).
Could you help me explaining the elements of $\operatorname{Gal}(L/K)$?