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.
You mean $L/K$ is Galois, $G = Gal(L/K)$, $H$ a normal subgroup, $L^H$ its fixed field.
Moreover $K= L^G$ (you can take $K = L^G$ as the definition of $L/K$ is Galois with Galois group $G$)
For $\sigma \in G$ then $\sigma(L^H) = L^{\sigma H \sigma^{-1}}$. As $H$ is normal then $\sigma H \sigma^{-1} = H$ so
$\sigma(L^H) =L^H$ and $\sigma |_{L^H} \in Gal(L^H/K)$.
The restriction to $L^H$ sends $G$ to $G/ \ker(\sigma \to \sigma|_{L^H}) = G/ Gal(L/L^H) = G/H$.
Thus it makes sense to look at the fixed subfield $(L^H)^{G/H} = L^G = K$ which implies $L^H/K$ is Galois with Galois group $G/H$.
If $L/K$ is not Galois then let $H = \{1\}$ to see it cannot work.
Best Answer
Huh, funny, we just went over this today in my algebra class.
Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.
Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,b\in H$, $ab=ba$ since it must also hold in $G$ (as $a,b \in G \ge H$ and $G$ is given to be abelian).