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I am asked to deduce, in the following order, 3 facts about $G = Gal(f(t) = t^6+3)$ over the rationals:
- $G$ is order 6
- The elements have orders $1,2,3$
- $G \cong S_3$
I have figured these out but in the wrong order so I wanted to write down a different train of thought that I am less sure of to see if there is anything wrong with it.
By Eisenstein $f(t)$ is irreducible, and is thus the minimal polynomial of a root $\beta$ of $f$ over $\mathbb Q$, so if $L$ is the splitting field of $f$ over $\mathbb Q$ then as the degree of $f$ is $6$, then the degree of the extension $\mathbb Q\leq L$ is also $6$.
Letting $\xi$ be a primitive $6$-th root of unity, then we see that over $\mathbb Q(\xi) = \mathbb Q(i\sqrt 3), \; f$ factorises as $f(t) = (t^3 + i\sqrt 3)(t^3 - i\sqrt 3)$ . This means that all permutations $\sigma \in Gal(L/\mathbb Q(\xi))$ permute the roots of each cubic factor amongst themselves. i.e. $\sigma$ is a $3$-cycle, and there are three of them. Also we note that complex conjugation preserves $\mathbb Q$ in $L$, so $G$ contains an order $2$ permutation. The identity is order $1$. I am not quite sure how to conclude that there is no element of order $6$, but with that we have that the elements are of orders $1,2,3$
Finally, we notice that composing complex conjugation with one of the $3$-cycles gives us the inverse of the original $3$-cycle. This gives a dihedral relation and hence $G$ is the dihedral group of order $6$, i.e. $G \cong S_3$
I would like to clarify why it is exactly that we can conclude that there is no element of order $6$ in $G$. I can feel that it is a very simple thing that I have overlooked, but I can't figure out what it is and would appreciate a little help or a hint. Additionally I would really appreciate it if anyone could let me know if there's anything flawed with my logic here at all. Thank you.