Galois group of $\mathbb Q(\sqrt{4+\sqrt 7})/\mathbb Q$ Let $\alpha =\sqrt{4+\sqrt 7}$ and $\beta =\sqrt{4-\sqrt 7 }$. I have to compute $Gal(\mathbb Q(\alpha )/\mathbb Q)$. So, I found that $Gal(\mathbb Q(\alpha )/\mathbb Q)=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$, that $\mathbb Q(\beta^2)= \mathbb Q(\alpha^2)$. I show that the roots of the minimal polynomial of $\alpha $ are $\{\alpha ,-\alpha ,\beta ,-\beta \}$. But now that ask me to enumerate the element of the galois group. I did it's $\{id, \sigma ,\tau,\sigma \tau\}$ where 
\begin{align*}
\sigma : \alpha &\longmapsto -\alpha \\
\beta &\longmapsto -\beta 
\end{align*}
and 
\begin{align*}
\tau : \alpha &\longmapsto \beta  \\
\beta &\longmapsto \alpha 
\end{align*}
But in my correction they wrote 
\begin{align*}
\sigma : \alpha &\longmapsto -\alpha \\
\beta &\longmapsto \beta 
\end{align*}
Question : Who is right ? me or the correction ?
The problem it's that this group is abelian, so if they are right, $$\sigma \tau(\alpha )=\beta \neq -\beta =\tau\sigma (\alpha )$$
so I think I'm right, but in the doubt I prefer to ask you.

Notice that
$$\alpha\beta =\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=\sqrt{16-7}=3,$$
so $\beta=\frac{3}{\alpha}$. As a consequence, any homomorphism $\varphi$ such that $\varphi(\alpha)=-\alpha$ is also such that$$\varphi(\beta)=\varphi\left(\frac{3}{\alpha}\right)=\frac{3}{\varphi(\alpha)}=\frac{3}{-\alpha}=-\beta.$$
So the error is in your correction. Your answer is right.
