Show that Galois Group $G(K/Q)$ of K over Q is isomorphic to symetric group $S_3$

Question

Let K be the splitting field of $f(X)=X^3-2$ over Q. where Q is the set of all the rational numbers

(a)determine an explicit set of generators for K over Q

(b)show that the Galois Group $G(K/Q)$ of K over Q is isomorphic to the symmetric group $S_3$.

Thus, I know, as for part(a) we have $K=Q(2^{1/3},\sigma)$, where $\sigma$ is the complex cube root of unit.Now ,we should have $|K:Q|=6$, but how can we prove that $G(K/Q)$ is isomorphic to $S_3$.

Answer 1

Hint: There are only two groups of order $6$ namely $S_3$ and cyclic group of order $6$,$G(K/Q)$ cannot be cyclic group because $Q(2^{1/3})/Q$ is not Galois.[Fundamental Theorem of Galois Extensions]

Answer 2

As you said, the splitting field is $\mathbb{K}=\mathbb{Q}(\sqrt[3]{2}, \zeta)$, where $\zeta$ is the primitive complex root of 1 given by $e^{(\frac{2i\pi}{3})}$. Let's now focus on two sub extensions of $\mathbb{K}$: $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\zeta)$. Their degrees over $\mathbb{Q}$ are respectively 3 and 2, for you can apply Eisenstein Criterion to f(x) with $p=3$ and see that $\mathbb{Q}(\zeta)$ is a cyclotomic extension of $\mathbb{Q}$ of degree $\phi(3)=2$, where $\phi$ stands for Euler's function. Thus we know that $[\mathbb{K}:\mathbb{Q}]=6$ for we have $[\mathbb{K}:\mathbb{Q}] \leq (deg(f(x))!$ and $[\mathbb{K}:\mathbb{Q}] \geq gcd(3,2)$. We now know that the Galois group is finite with 6 elements.

$\mathbb{Q}(\zeta)$ is a cyclotomic extension of $\mathbb{Q}$$\Rightarrow$ it is a normal sub extension for it is indeed the splitting field of the third cyclotomic polynomial $\Rightarrow$ the Galois group has a normal subgroup of order $\frac{6}{2}=3$ (Galois' second theorem). We also have a subgroup of order 2 given by the conjugate map $\sigma \in Gal(\mathbb{K/Q})$ that fixes $\sqrt[3]{2}$ and maps $\zeta$ onto $\zeta^{2}$ - which is precisely its complex conjugate.

Let $\tau \in Gal(\mathbb{K/Q})$ be the automorphism that fixes $\zeta$ and maps $\sqrt[3]{2}$ to $\sqrt[3]{2}\zeta$. You can easily check that $\sigma\tau\sigma=\tau^{-1}$, which makes $<\tau,\sigma> \cong D_{3} \cong S_{3}$, where by $D_{3}$ I mean the Dihedral group with 6 elements.

Since both $\tau$ and $\sigma$ are in $Gal(\mathbb{K/Q})$, we have that $Gal(\mathbb{K/Q}) \cong S_{3}$.