Show $Gal(E/\mathbb{Q})\cong \mathbb{\mathbb{Z_2\times Z_2}}$ Let $ f(x) = x^4 −4 ∈ \mathbb{Q}[x],$ find the splitting field $E$ of $f$ over $\mathbb{Q}$, then show that $Gal(E/\mathbb{Q})\cong \mathbb{\mathbb{Z_2\times Z_2}}$

the roots of $f$ are $\sqrt2,-\sqrt2,i\sqrt2,-i\sqrt2$ so the spliting field $E=\mathbb{Q}(\sqrt2,i\sqrt2)$
the elements of  $Gal(E/\mathbb{Q})$ are:
$σ_0=id, σ_1(\sqrt2)=-\sqrt2, σ_2(i\sqrt2)=-i\sqrt2$ , but these are $3$ elements how it's possible $Gal(E/\mathbb{Q})\cong \mathbb{\mathbb{Z_2\times Z_2}}$?
Where is my mistake?
Edit after koljas comment, it should be $σ_0=id, σ_1:\sqrt2\rightarrow -\sqrt2 \:\text{and}\: i\sqrt2\rightarrow i\sqrt2, σ_2: \sqrt2\rightarrow \sqrt2 \:\text{and}\: i\sqrt2\rightarrow -i\sqrt2 , σ_3: \sqrt2\rightarrow -\sqrt2 \:\text{and}\: i\sqrt2\rightarrow -i\sqrt2 $
 A: I want to give a thourough explanation of what is going on. Our goal ist to understand $Gal(E/\mathbb{Q})$ where $E$ is the splitting field of $X^4-4\in \mathbb{Q}[X]$. Clearly in $\mathbb{C}[X]$ this polynomial factors $(X^2-2)(X^2+2)=(X-\sqrt{2})(X+\sqrt{2})(X+i\sqrt{2})(X-i\sqrt{2})$ and the splitting field is thus genereated by the roots of this polynomial, hence $E=\mathbb{Q}(\sqrt{2},i\sqrt{2})$. We know that $\#Gal(E/\mathbb{Q})=[E:\mathbb{Q}]=4$, because we have $[E:\mathbb{Q}]=[E:\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2\cdot 2=4$.
Consider the canonical embedding $\sigma:\mathbb{Q}\to \mathbb{Q}(\sqrt{2})$ by a standard theorem we know that for every root of $X^2-2$ in $\mathbb{Q}(\sqrt{2})$ we get a continuation of $\sigma$ that send $\sqrt{2}$ to the root. Let $\sigma':\mathbb{Q}(\sqrt{2})\to \mathbb{Q}(\sqrt{2})\hookrightarrow E$ be the canonical continuation of such a continuation to $E$ by inclusion. Apllying the same argument again we obtain the four elements of the galois group:
$$ \sigma_1=\operatorname{id}_E; \ \sigma_2:\sqrt{2}\to -\sqrt{2},i\sqrt{2}\to i\sqrt{2};\ \sigma_3:\sqrt{2}\to \sqrt{2},i\sqrt{2}\to -i\sqrt{2}; \ \sigma_4:\sqrt{2}\to -\sqrt{2},i\sqrt{2}\to -i\sqrt{2}$$
In particular we know that $Gal(E/\mathbb{Q})$ is $\cong \mathbb{Z}/4\mathbb{Z}$ or $\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. In order to find out which of those is the case we look at the order of the elements in $Gal(E/\mathbb{Q})$, but one easily sees that no element has order $4$. Hence, $Gal(E/\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.
