# 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

$$\langle \sigma_2, \sigma_3, \sigma_5 \rangle \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}}$$

• Since quotients are built from left cosets, we write them as $$G/H$$ and not $\frac{G}{H}$.
– Shaun
Commented Apr 3, 2022 at 12:17
• @Shaun I don't think we ever really write it as $\frac GH$. Left, right or two-sided cosets have always been elements of $G/H$ that I've seen. Commented Apr 3, 2022 at 12:25

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