I need help for the following problem(what is the key-idea that problem):
Problem: Let $f(x)$ be a monic polynomial with rational coefficients. Assume $f(x)$ is irreducible in $\mathbb{Q}[x]$ and the Galois-group of $f(x)$ over $\mathbb{Q}$ is a group of order 99. What is the degree of $f(x)$?
Solution(my attempt): Let $\alpha$ be a root of $f \Longrightarrow \big[ \mathbb{Q}(\alpha) : \mathbb{Q} \big] = deg(f)$.
Let $K$ be the splitting field of $f$ over $\mathbb{Q} \Longrightarrow \Big|Aut_{\mathbb{Q}}(K) \Big| = 99$.
How can we compute the degree of the polynomial $f$?
Note that a group of order $99=3^2\times 11$ is abelian: by Sylow's Theorems, there is a unique Sylow $11$-subgroup and a unique Sylow $3$-subgroup, and both are necessarily abelian by order considerations.
The degree of $f$ must divide the order of the Galois group. In addition, because the Galois group permutes the roots of $f$ and is completely determined by the action on those roots, it must be a transitive abelian subgroup of $S_n$, where $\deg(f)=n$.
So we know that $n$ divides $99 =3^2\times 11$. And that $99$ divides $n!$.
However, transitive abelian subgroups of $S_n$ have order $n$, so that would seem to require that $n=99$.
Now, does this make sense? Yes!
The degree can definitely be $99$, since by the Primitive Element Theorem the extension is simple, and if the extension is $\mathbb{Q}(a)$, then the minimal polynomial of $a$ is monic, irreducible, and of degree $99$. So from a purely "meta" perspective, if this answer is going to have a definite answer, that answer must be $99$... and the fact about abelian transitive subgroups seals the deal.