Explicit expression of Galois group of $x^8-5$ over $\mathbb{Q}$

Question

Let $F$ be the splitting field of $x^8-5$, we have $F=\mathbb{Q}(5^{1/8},i,\sqrt{2})$. I need to calculate the galois group Gal$(F/\mathbb{Q})$.

I know the three generators are \begin{align*} &\sigma_1:5^{1/8}\mapsto 5^{1/8}\zeta_8,\;\sqrt{2}\mapsto\sqrt{2},\;i\mapsto i\\ &\sigma_2:5^{1/8}\mapsto 5^{1/8},\;\sqrt{2}\mapsto-\sqrt{2},\;i\mapsto i\\ &\sigma_3:5^{1/8}\mapsto 5^{1/8},\;\sqrt{2}\mapsto\sqrt{2},\;i\mapsto -i\\ \end{align*}

Then Gal$(F/\mathbb{Q})=<\sigma_1,\sigma_2,\sigma_3>$

So I got (updated: adding $\beta\gamma=\gamma\beta$) $$ Gal(F/\mathbb{Q})=<\alpha,\beta,\gamma:\alpha^8=1,\beta^2=\gamma^2=1,\gamma\beta=\beta\gamma, \beta\alpha=\alpha^3\beta,\gamma\alpha=\alpha^7\gamma> $$

But I am not sure does such result count as working out the galois group? Is there any way to work out an explicit expression like $D_8\times K$ or something?

I know $|Gal(F/\mathbb{Q})|=32$, and $<\sigma_2,\sigma_3>=\mathbb{Z}_2\times\mathbb{Z}_2$, and $<\sigma_1>=\mathbb{Z}_8$, but how to combine them?

Answer 1

Let $G=\mathbb{Z}_8\rtimes \mathbb{Z}_8^\times$. Elements of $G$ are pairs $(a,b)$ with $a\in\mathbb{Z}_8$ and $b\in\mathbb{Z}_8^\times$. The group operation is given by $(a,b)(a',b')=(a+ba',bb')$. Since $(a,b)=(a,1)(0,b)$, it is generated by elements of the form $(a,1)$ and $(0,b)$. Note that the inclusion maps $i_1:\mathbb{Z}_8\rightarrow G$ and $i_2:\mathbb{Z}_8^\times\rightarrow G$ given by $i_1(a)=(a,1)$ and $i_2(b)=(0,b)$ are homomorphisms. Since $\mathbb{Z}_8^\times$ is generated by 3 and 7, we can use the above facts to conclude that $G$ is generated by the elements $\alpha=(1,1)$, $\beta=(0,3)$ and $\gamma=(0,7)$.

Since $i_1$ is a homomorphism, $\alpha^8=(0,1)$. Since $i_2$ is a homomorphism, $\beta^2=\gamma^2=1$ and $\beta\gamma=\gamma\beta$. Furthermore, $\beta\alpha=(0,3)(1,1)=(3,3)$ and $\alpha^3\beta=(1,1)^3(0,3)=(3,1)(0,3)=(3,3)$ so $\beta\alpha=\alpha^3\beta$. Similarly, $\alpha^7\gamma=\gamma\alpha$. Using the fact that $G$ has order 32, one may show these give all the relations. By comparing with your presentation for $\mathrm{Gal}(F/\mathbb{Q})$, we see that $G\cong \mathrm{Gal}(F/\mathbb{Q})$.

Alternately, we could have made this identification directly by using the fact that $G$ acts on $F$ by $(a,b) \zeta_8=\zeta_8^b$ and $(a,b)5^{1/8}=\zeta_8^a 5^{1/8}$.

We can give a slightly different description as follows. Let $H$ be the subgroup of $GL_2(\mathbb{Z}_8)$ consisting of matrices of the form $ \left[ \begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$ with $a\in \mathbb{Z}_8$ and $b\in\mathbb{Z}_8^\times$. Define $\phi:G\rightarrow H$ by $\phi(a,b)= \left[\begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$. One may check that $\phi$ is an isomorphism, so the Galois group can be identified with the group $H$ of invertible matrices over $\mathbb{Z}_8$ of the form $ \left[ \begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$.