# Associating transitive subgroups of $S_5$ to field extensions

Problem 5.3 from Morandi: Field and Galois Theory Find all the transitive subgroups of $$S_5$$ whose order is a multiple of 5, and for each, find a field $$F$$ and an irreducible $$f$$ over $$F$$ such that $$\text{Gal}(f/F)$$ is isomorphic to the given subgroup. (Hint: this will require semidirect products)

Solution Attempt: By brute force, $$\langle(12345)\rangle$$ is transitive, which also means that $$A_5\cong\langle(12345),(123)\rangle$$ and $$D_{10}\cong\langle(12345),(25)(34) \rangle$$ and $$\langle(12345),(2354)\rangle$$ are transitive. So there are five transitive subgroups up to conjugacy, all of which have orders a multiple of 5 (I'm pretty sure any transitive subgroup has an order a multiple of 5 by the orbit-stabilizer theorem).

Anyway, here's where I got a little stuck, as I'm not sure why I would need semidirect products. I've just been trying random polynomials to get field extensions of the right degree, and I know there must be a more systematic way.

I've been able to realize $$S_5$$ as the Galois group of $$x^5-6x+3$$. It's an irreducible over $$\mathbb{Q}$$, and I know it has three real roots and two complex roots. Complex conjugation is a $$\mathbb{Q}$$-automorphism of order 2, so our Galois group can't be $$A_5$$ since $$A_5$$ doesn't contain any transpositions. Suppose $$\sigma$$ is our transposition. There is also an element of order 5, $$\tau$$ in the Galois group, and so we can make an element of order 3 by taking $$(\sigma \tau^2)^2$$, and this ensures out Galois group is none of the other transitive subgroups of $$S_5$$ because 3 doesn't divide their orders.

We can realize $$\langle(12345),(2435)$$ as the Galois group of the splitting field of $$x^5-2$$ over $$\mathbb{Q}$$. $$x^5-2$$ splits over $$\mathbb{Q}(e^{2\pi i}{5},\sqrt{2})$$ and $$[\mathbb{Q}(e^{2\pi i}{5},\sqrt{2}):\mathbb{Q}]=20.$$ I guess this has to mean this one is isomorphic to $$\mathbb{Z}_5\rtimes \mathbb{Z}_4$$

I'm having trouble with the rest of them. I think I can get $$A_5$$, but I can't think of field extensions of degree 5 or 10.

The transitive subgroups of $$S_5$$ are isomorphic to $$S_5,A_5,\mathbb{Z}/5\mathbb{Z},D_5$$ and $$\mathbb{Z}/5\rtimes (\mathbb{Z}/5)^\times$$.
You got $$S_5$$ and the last case. For the cyclic group of order $$5$$ observe that $$\mathbb{Q}(\zeta_{11})/\mathbb{Q}$$ is Galois, with Galois group $$(\mathbb{Z}/11)^\times\simeq \mathbb{Z}/10$$. It is easy to deduce that $$\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q}$$ is Galois, with Galois group $$\mathbb{Z}/5\mathbb{Z}$$. For $$A_5$$ and $$D_5$$ this is more delicate. Since i'm a bit lazy today, i let you read the following paper: https://rak.ac/files/papers/galois.pdf