## Tag Archives: dihedral group

### Galois group of order 6

$\mathbf{Problem:}$ Let $E/\mathbb{C}(t)$ where $t$ is transcedental over $\mathbb{C}$ and let $w\in \mathbb{C}$ satisfy $\displaystyle w^3=1$, $w\neq 1$. Let $\sigma$ be the automorphism of $E/\mathbb{C}$ such that $\tau(t)=t^{-1}$. Show that

$\displaystyle \sigma^3=1=\tau^2,\quad \tau\sigma=\sigma^{-1}\tau$.

Show that the group of automorphism generated by $\sigma$ and $\tau$ has order 6 and the subfield $F=\text{Inv}G=\mathbb{C}(u)$ where $u=t^3+t^{-3}$.

$\mathbf{Proof:}$ It is easy to verify the structure of group $G$. And $G$ is the dihedral group $D_6$

Let $H_1=\langle\sigma\rangle\leq G$, $H_2=\langle\tau\rangle\leq G$. Then $\displaystyle \text{Inv}H_1=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(\omega t)}{g(\omega t)}\right\}=\left\{\frac{f(t^3)}{g(t^3)}|g(t) \neq 0\right\}$.

Consider $\displaystyle \text{Inv}H_2=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(t^{-1})}{g(t^{-1})}\right\}$, any $\displaystyle \frac{f(t)}{g(t)}\in \text{Inv}H_2$$\exists\, h, l\in \mathbb{C}(t)$ such that

$\displaystyle 2\frac{f(t)}{g(t)}=\frac{f(t)}{g(t)}+\frac{f(t^{-1})}{g(t^{-1})}=\frac{f(t)g(t^{-1})+g(t)f(t^{-1})}{g(t)g(t^{-1})}=\frac{h(t+t^{-1})}{l(t+t^{-1})}$

So $\displaystyle \text{Inv}H_2=\left\{\frac{f(t+t^{-1})}{g(t+t^{-1})}\lvert g\neq 0\right\}$.

So $\text{Inv}G=\text{Inv}H_1\cap\text{Inv}H_2=\mathbb{C}(t^3+t^{-3})$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Jacobson p243