Deducing Galois group of $t^6 + 3$ over the rationals I am asked to deduce, in the following order, 3 facts about $G = Gal(f(t) = t^6+3)$ over the rationals:


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*$G$ 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.


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*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. 
 A: You have basically the right idea but there are some details you haven't stated exactly right.  Here is how I would clean up your argument.
First, in your step 1, you have only found that $[K:\mathbb{Q}]=6$,  not that $[L:\mathbb{Q}]=6$.  This tells you $|G|=[L:\mathbb{Q}]\geq 6$.  Equality does follow from some of your later work though.
Let's now look at step 2.  To justify your assertions about $Gal(L/\mathbb Q(\xi))$ more precisely, let's let $\alpha$ be a chosen cube root of $i\sqrt{3}$.  Then from your factorization of $f$, we see that the roots of $f$ are $\alpha$, $\alpha\xi^2$, $\alpha\xi^4$, $-\alpha$, $-\alpha\xi^2$, and $-\alpha\xi^4$.  This proves that $L=\mathbb{Q}(\alpha,\xi)=\mathbb{Q}(\alpha)$ since $\xi\in\mathbb{Q}(\alpha^3)$, and so proves that in the previous paragraph, $K=L$ and $G$ really does have only $6$ elements.  We also see that $[L:\mathbb{Q}(\xi)]=[L:\mathbb{Q}]/[\mathbb{Q}(\xi):\mathbb{Q}]=3$, so $Gal(L/\mathbb Q(\xi))$ can only be a cyclic group of order $3$.  Since $L$ is the splitting field of $t^3-i\sqrt{3}$ over $\mathbb{Q}(\xi)$, this cyclic group of order $3$ must permute the cube roots of $i\sqrt{3}$ cyclically, as you claimed.  (But there are only two such elements of order $3$; the third element is the identity!)
So at this point, we have found two elements of order $3$ and one element of order $1$ in $G$ (forming a cyclic subgroup of order $3$).  You then correctly observe that complex conjugation is in $G$, giving an element of order $2$.
At this point I don't see any direct way to conclude $G$ has no elements of order $6$.  However, if $G$ had an element of order $6$, it would be cyclic, so it suffices to show $G$ is not abelian.  So, let's look at your step 3 now.  You claim that composing complex conjugation with one of the 3-cycles gives its inverse, but that is false!  Rather, conjugating one of the 3-cycles by complex conjugation gives its inverse.
To show this, let's write everything down explicitly in terms of $\alpha$ and $\xi$.  There is a generator $\sigma\in Gal(L/\mathbb{Q}(\xi))$ such that $\sigma(\alpha)=\alpha\xi^2$.  If $\tau$ denotes complex conjugation, we have $\tau(\xi)=\xi^{-1}$.  To find $\tau(\alpha)$, note first that $\tau(\alpha)^3=\tau(i\sqrt{3})=-i\sqrt{3}$, so $\tau(\alpha)$ is one of $-\alpha$, $-\alpha\xi^2$, and $-\alpha\xi^4$.  Note also that $\alpha$ is not purely imaginary, so $\tau(\alpha)$ cannot be $-\alpha$.  We thus may assume $\tau(\alpha)=-\alpha\xi^2$ (if not, then change which primitive 6th root we're calling $\xi$ to make it true).
Now we can compute the compositions of $\sigma$ and $\tau$ explicitly.  We have $$\sigma(\tau(\alpha))=\sigma(-\alpha\xi^2)=-\sigma(\alpha)\xi^2=-\alpha\xi^4$$ and $$\tau(\sigma(\alpha))=\tau(\alpha\xi^2)=-\alpha\xi^2\xi^{-2}=-\alpha.$$
In particular, $\sigma\tau\neq\tau\sigma$, so $G$ is nonabelian and has no elements of order $6$.  At this point you can conclude $G\cong S_3$ since $S_3$ is the unique nonabelian group of order $6$ up to isomorphism.  Alternatively, you can use similar computations to those above to find that in fact $\tau\sigma\tau=\sigma^{-1}$ and get the dihedral relation you referred to and an explicit isomorphism to $S_3$.
