# Unconventional Galois theory in Qual

This seems to be an uncommon question that one may see in Qual Algebra exams, and I don't think if it was a wise choice to put this kind of questions in Qual. I consider it as a hard Galois theory question, though!

Let $f(x) \in \mathbb{Q}[x]$ be a polynomial of degree $7$ and let $E$ be the splitting field of $f$ over $\mathbb{Q}.$ Assume that $\text{Gal}(E/\mathbb{Q}) \cong S_7.$

a) Find the number of intermediate fields $K$ between $\mathbb{Q}$ and $E$ such that $[E:K]=9.$

b) Show that the intersection of all fields $K$ in part a) is not equal to $\mathbb{Q}.$

c) If $\alpha \in E$ is a root of $f(x),$ how many of the intermediate fields in part a) contain $\alpha?$

• a) Is just group theory, find all subgroups of order $9$. b) Is also group theory, show that the subgroup generated by all those subgroups is not $S_7$. c) After you've figured out the form of the subgroups in (a) this one should be clear based on which ones fix a root. – JSchlather Jan 8 '13 at 22:17

a) As I noted in my comment this problem reduces to finding all subgroups of order $9$ of $S_7$. These are the sylow subgroups. It's not hard to see that any two disjoint $3$-cycles will generate such a subgroup. Furthermore since all of these subgroups are conjugate and the action by conjugation in $S_n$ by an element $\sigma$ is given by $\sigma(123)\sigma^{-1}=(\sigma(1)\sigma(2)\sigma(3))$ it follows that all such Sylow subgroups are generated by two disjoint $3$-cycles. Now to count them note that any choice of $3$ elements determines a $3$-cycle so we have $\binom{7}{3}$ for the first $3$ cycle and $\binom{4}{3}$ for the second $3$-cycle. Accounting for double counting leaves us with $70$ such subgroups.
b) Note that every Sylow $3$-subgroup is contained in $A_7$. It follows that the intersection of all the fields in part a) contains the fixed field of $A_7$.
c) Now every such subgroup was generated by two disjoint $3$-cycles so left one element or rather root fixed. So we want to know how many of these subgroups fixed one root then we have $6$ elements and $\binom{6}{3}=10$ ways to pick a $3$-cycle which determines the other one.