As John Martin pointed out, you can take any root $\alpha \in K$ of $f$ and then you show that $\Bbb Q(\alpha)$ is Galois over $\Bbb Q$.
In particular, this means that every root of the minimal polynomial of $\alpha$ over $\Bbb Q$ is contained in $\Bbb Q(\alpha)$. Since $f$ is irreducible, the minimal polynomial of $\alpha$ over $\Bbb Q$ is actually $f$. Therefore, $\Bbb Q(\alpha)$ is the splitting field of $f$, which is $K$.
Why is $\Bbb Q(\alpha)$ is Galois over $\Bbb Q$ ? Well, you know that if $L/F$ is a Galois extension, then there is a correspondence between :
- the subextensions $M/F$ of $L$ such that $M/F$ is Galois
- the normal subgroups of $G = \text{Gal}(L/F)$
(If $H \trianglelefteq G\,$, then $\text{Gal}(L^H/F) = G/H\,$, where $L^H = \{ x \in L \;\mid\; \sigma(x)=x, \;\forall \sigma \in H \}$).
In our case, every subgroup of $\text{Gal}(K/\Bbb Q)$ is normal, so that every subextension $M/\Bbb Q$ of $K$ is Galois over $\Bbb Q$.
Here are some details about the correspondence.
Suppose that $F \subset M \subset L\;$ are field extensions, such that $L/F$ is Galois with abelian Galois group.
You want to prove that $M/F$ is Galois. It is not difficult to show that it is separable.
To show that it is a normal extension, let $\sigma \in \text{Gal}(L/F)$, $m \in M\;$ and let's show that $\sigma(m) \in M$.
Since $L/M$ is Galois, we know that $M=L^{\text{Gal}(L/M)}\;$, so it is sufficient to show that $\sigma(m)$ is fixed by every $\tau \in \text{Gal}(L/M) \subset \text{Gal}(L/F).\;$
You have $\tau(\sigma(m)) = \sigma(\tau(m)) = \sigma(m),\;$ because $\text{Gal}(L/F)$ is abelian and $\tau$ fixes $M$. QED.
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.
Best Answer
By the Galois correspondence, there is an intermediate field $E$ with $L/E$ of degree $p$ (hence $E/K$ of degree $q$), corresponding to a subgroup of order $p$ of $G(L/K)$. By the Galois correspondence, $E$ is not Galois over $K$, since the subgroups of order $p$ are not normal in $G(L/K)$. But $E$ is a simple extension over $K$, so $E=K(\rho)$ for some $\rho\in E\setminus K$. This $\rho$ is the root of an irreducible polynomial in $K[x]$ of degree $q$, and its splitting field is strictly larger than $E$ (as otherwise $E$ would be a splitting field over $K$ and hence Galois over $K$), and contained in $L$ (because $L$ is Galois).