Question

Write out the operation table for the group $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathbf{}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)}\mathbf{=}\left\{i,\tau .\alpha ,\beta \right\}$.

Short answer

The operation table is:

$$\begin{gathered} \left[\mathit{i}\left(\sqrt{3},\sqrt{3}\right)\mathbf{,}\mathit{\tau }\left(\sqrt{3},-\sqrt{3}\right)\mathbf{,}\mathit{\alpha }\left(\sqrt{3},\sqrt{3}\right)\mathbf{,}\mathit{\beta }\left(\sqrt{3},-\sqrt{3}\right)\right] \\ \left[\mathit{i}\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\tau }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{-}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\alpha }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\beta }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{-}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\right] \end{gathered}$$

Step-by-step solution

Definition of Galois group

The set of all F automorphisms of K is an extension of F then Galois group is denoted by$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$ . It is the Galois group K of over F.

Operation table for the group.

Consider the Galois group of Q$\left(\sqrt{3},\sqrt{5}\right)$over the Q. There are two minimal polynomial equations are ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$and ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}$. The above expression implies that it takes the roots $\sqrt{\mathbf{3}}$to $\sqrt{\mathbf{3}}$or $-\sqrt{\mathbf{3}}$and to $\sqrt{\mathbf{5}}$or $-\sqrt{\mathbf{5}}$.

Since, there are at most four automorphism in given $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\mathbf{}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)}$over Q .

So, the possibilities are:

$\sqrt{\mathbf{3}}\stackrel{\mathbf{i}}{\mathbf{\to }}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{\tau }}{\mathbf{\to }}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{\alpha }}{\mathbf{\to }}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{\beta }}{\mathbf{\to }}\mathbf{-}\sqrt{\mathbf{3}}$

Same as:

$\sqrt{\mathbf{5}}\stackrel{\mathbf{i}}{\mathbf{\to }}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\stackrel{\mathbf{\tau }}{\mathbf{\to }}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\stackrel{\mathbf{\alpha }}{\mathbf{\to }}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\stackrel{\mathbf{\beta }}{\mathbf{\to }}\mathbf{-}\sqrt{\mathbf{5}}$

Now, the given function is a group of order 4 by constructing nonidentity automorphism localid="1657967199712" $\mathit{(}\mathit{r}\mathit{,}\mathit{\alpha }\mathit{,}\mathit{\beta }\mathit{)}$.

Now, the minimal polynomial is isomorphism.

So,

$\mathbf{\sigma }\mathbf{:}\mathbf{}\mathbf{Q}\left(\sqrt{3}\right)\mathbf{\cong }\mathbf{Q}\left(-\sqrt{3}\right)$

$\mathit{\sigma }\left(\sqrt{3}\right)\mathbf{=}\mathbf{-}\sqrt{\mathbf{3}}$

For ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}$over $\mathbf{Q}\left(\sqrt{3}\right)$and $\mathrm{\sigma }\to \mathrm{Q}$automorphism of

$\mathit{Q}\left(\sqrt{3}\right)\left(\sqrt{5}\right)\mathbf{=}\mathit{Q}\left(\sqrt{3}\right)\left(\sqrt{5}\right)$.Since,$\mathit{r}\left(\sqrt{5}\right)\mathbf{=}\sqrt{\mathbf{5}}$.Therefore,$\mathit{r}\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\left(\sqrt{3},\sqrt{5}\right)$

Now,

$$\begin{gathered} \mathit{r}\left(\sqrt{3}\right)\mathbf{=}\mathbf{-}\mathit{\sigma }\mathbf{(}\sqrt{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\sqrt{\mathbf{5}} \end{gathered}$$

Further each$\left(r,\alpha ,\beta \right)$has order 2 in$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\left(\sqrt{3},\sqrt{5}\right)$.

Then,

localid="1657968372971" $$\begin{gathered} \mathit{\tau }\mathbf{\circ }\mathit{\tau }\mathbf{\left(}\sqrt{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathit{\tau }\mathbf{\left(}\mathbf{\tau }\left(\sqrt{3}\right)\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{\tau }\left(-\sqrt{3}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathit{\tau }\left(\sqrt{3}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{\mathbf{3}} \end{gathered}$$

That is equal to $\mathit{i}\left(\sqrt{3}\right)$.

Since

$$\begin{gathered} \left(\tau \circ \tau \right)\left(\sqrt{3}\right)\mathbf{=}\sqrt{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{i}\left(\sqrt{3}\right) \end{gathered}$$

Similarly,

$$\begin{gathered} \mathbf{(}\mathbf{\tau }\mathbf{\circ }\mathbf{\tau }\mathbf{)}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{5}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{i}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)} \end{gathered}$$

Hence, operator $\left(\tau \circ \tau \right)\mathbf{=}\mathit{i}$and $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\left(\sqrt{3},\sqrt{5}\right)\mathbf{\cong }{\mathit{Z}}_{\mathbf{2}}\mathbf{\times }{\mathit{Z}}_{\mathbf{2}}$. Also $\left(\tau \circ \alpha \right)\mathbf{=}\mathit{\beta }$. Since, the operation table is:

$$\begin{gathered} \left[\mathit{i}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\tau }\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\alpha }\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\beta }\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{}\mathbf{)}\right] \\ \left[\mathit{i}\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\tau }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{-}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\alpha }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\mathbf{,}\mathit{\beta }\mathbf{(}\sqrt{\mathbf{5}}\mathbf{-}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{}\mathbf{)}\right] \end{gathered}$$