why not $G \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} ?$

Question

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}} $

Answer 1

$\sigma_2$ 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 $$ G=\langle \sigma_2, \sigma_3, \sigma_5 \rangle \cong {\mathbb{Z}}/{2\mathbb{Z}} \times {\mathbb{Z}}/{2\mathbb{Z}} \times {\mathbb{Z}}/{2\mathbb{Z}} $$