Consider the polynomial $ f=X^{3}-2tX+t $ over $ K=\mathbb{C}(t) $, where $ t $ is transcendental over $ \mathbb{C} $. Let $ \alpha $ be a root of $ f $ and $ L $ be the splitting field of $ f $. I proved that $ K(\alpha) \subsetneq L=K(\alpha, \beta) $, where $ \beta $ is a root of the polynomial $ X^{2}+3\alpha^{2}-8t \in K(\alpha)[X] $. So the degree $ [L:K] $ must be 6.
I am asked to show further that $ Gal(L/K) \simeq S_{3} $ and to determine a a polynomial $ g \in K[X] $ such that $ M $ is the splitting field of $ g $, where $ M $ is the unique intermediate field of the extension $ L/K $ of degree 2. Additionally, is $ M/K $ Galois? From what I proved so far, I know that the Galois group of $ f $ is either $ \mathbb{Z}/6\mathbb{Z} $ or $ S_{3} $, but how do I show that it can't be abelian? Also, I dont know how to answer the other questions.
I would appreciate any help. Thank you!
The galois group permutes the roots of the polynomial, and an element is completely determined the images of the roots, so it must be a subgroup of $S_3$, as it has order $6$, it is the whole of $S_3$. Alternatively, if the extension was abelian, every subgroup the Galois group would be normal, hence by Galois correspondence, every subextension would be Galois, but from what you computed, $K(\alpha)$ is not Galois as it does not contain all roots of $f$.