Let the roots be $a,b,c,d,e$ with $a,b $ non-real. Your subgroup is transitive and has at least one transposition $(a b)$. Since it is transitive you can conjugate your transposition by some element to get a transposition which moves $c$ (conjugation preserves the cycle structure), and likewise one which moves $d$ or $e$.

Case 1. If this is $(bc)$ or $(ac)$ you can find your 3-cycle as a product of transpositions.

Case 2. If it is $(cd)$ then you can use conjugation to find a transposition which moves $e$ and whichever one you get, you can find your 3-cycle by composing it with either $(ab)$ or $(cd)$.

You were almost finished with the calculation of $L$.

Let $\alpha = \sqrt[3]{2 + \sqrt{3}}$ and $\beta = \sqrt[3]{2 - \sqrt{3}}$.

You've already observed that $\mathbb{Q}(\alpha^3) = \mathbb{Q}(\sqrt{3})$. And since $\alpha$ is a cube root of somethign in $\mathbb{Q}(\alpha^3)$ (that is not already a cube), we must also have $\omega \in L$, and so $K = \mathbb{Q}(i,\omega) \subseteq L$ (where $\omega$ is a primitive cube root of unity)

Finally, $K(\alpha)/K$ is an abelian extension of degree 3, and the same goes for $K(\beta) / K$ -- the only question is whether $\beta \in K(\alpha)$.

Well, this can be answered by looking at $K^*$. In this particular case, $\alpha^3$ is actually a unit of $\mathcal{O}$, the ring of integers in $K$. Furthermore, we know its unit group is $\mathbb{Z} \times \mu$, where $\mu$ is the roots of unity. Since $\alpha^3$ is not a root of unity, it generates a subgroup of infinite order.

The same goes for $\beta^3$. And therefore $\alpha^3$ and $\beta^3$ have to be related -- and now something that we probably should have noticed right from the beginning becomes the obvious thing to check for: we actually have $\alpha^3 \beta^3 = 1$. In fact, $\alpha \beta = 1$ in $L$.

So now it's clear: $L = K(\alpha) = \mathbb{Q}(i, \omega, \sqrt[3]{2 + \sqrt{3}})$. The six roots of your polynomial are:

$$ \alpha^{i} \omega^j $$

where $i \in \{-1,1\}$ and $j \in \{0,1,2\}$.

As for the Galois group, we've seen that it has a quotient isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}/2$. It's also clear it has a copy of $S_3$ as a subgroup, as that's the Galois group of $L / \mathbb{Q}(\sqrt{3})$. If I was up to par on my group theory, that would probably be enough for me to identify the group. (peeking at the other answer shows my first wild guess, $D_6$, would have been correct)

## Best Answer

Let $N$ and $C$ be the normalizer and centralizer of $H$ in $S_p$, respectively. Then $G$ is a subgroup of $N$ since $H$ is normal in $G$. Therefore we can establish your result by showing $N/H$ is a cyclic group whose order divides $p-1$.

Since $H$ is cyclic of order $p$, it follows that $C = H$ and that $\text{Aut}(H)$ is cyclic of order $p-1$. The result now follows since $N/C$ is isomorphic to a subgroup of $\text{Aut}(H)$, which is a general property of normalizers and centralizers.