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 $\mathbb{Q}(e^{2\pi i}{5},\sqrt[5]{2})$ and $$[\mathbb{Q}(e^{2\pi i}{5},\sqrt[5]{2}):\mathbb{Q}]=20.$$ 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