<s>[INST] why not $G \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} ?$ 
Determine  the Galois  group   of $(x^2-2) (x^2-3)(x^2-5)$. Determine  all the subfields  of the splitting field  of this  polynomial.

My attempt: I found the answer  here
$G= Gal(K/\mathbb{Q})$ where$ K= \mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt  5)$
It is written that

$\sigma_2 \begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$


$\sigma_3: \begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$


$\sigma_5: \begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}$

then obviously  $G= \langle \sigma_2, \sigma_3, \sigma_4 \rangle  \cong \frac{\mathbb{Z}}{2\mathbb{Z}}  \times  \frac{\mathbb{Z}}{2\mathbb{Z}} \times  \frac{\mathbb{Z}}{2\mathbb{Z}}$
My question:   why $G=\langle \sigma_2, \sigma_3, \sigma_4 \rangle  \cong \frac{\mathbb{Z}}{2\mathbb{Z}}  \times  \frac{\mathbb{Z}}{2\mathbb{Z}} \times  \frac{\mathbb{Z}}{2\mathbb{Z}} ?$
Why not $\langle \sigma_2, \sigma_3, \sigma_5 \rangle  \cong \frac{\mathbb{Z}}{3\mathbb{Z}}  \times  \frac{\mathbb{Z}}{3\mathbb {Z}} \times  \frac{\mathbb{Z}}{3\mathbb{Z}} ?$
My  thinking: $\sigma_2$ has $3$ choice  similarly $\sigma_3 $ and $\sigma_5$  have $3$ choice   so i think
$\langle \sigma_2, \sigma_3, \sigma_5 \rangle  \cong \frac{\mathbb{Z}}{3\mathbb{Z}}  \times  \frac{\mathbb{Z}}{3\mathbb {Z}} \times  \frac{\mathbb{Z}}{3\mathbb{Z}} $
 [/INST] $\sigma_2$ doesn't have "three choices" it has 8:
$$
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
$$
Of course, to those of us in the know, only one of these deserve the name $\sigma_2$.
No, more important than the number of possibilities is its order. Choose
$$
\sigma_2:\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$$
Then $\sigma_2^2$ is the identity. This means that $\langle \sigma_2\rangle\cong \Bbb Z/2\Bbb Z$. Same for the other two $\sigma$'s. The way the three chosen $\sigma$'s interact (they commute, their generated subgroups intersect trivially, and together they generate the whole group) is what yields
$$
G=\langle \sigma_2, \sigma_3, \sigma_5 \rangle  \cong {\mathbb{Z}}/{2\mathbb{Z}}  \times  {\mathbb{Z}}/{2\mathbb{Z}} \times  {\mathbb{Z}}/{2\mathbb{Z}}
$$
</s>