If Gal(K,Q) is abelian then |Gal(K,Q)|=n

I wonder if this works out:

Since $K/ℚ$ is abelian, every intermediate extension is normal and so is $ℚ(α)$ for some zero $α ∈ K$ of $f$. This must mean that $ℚ(α)$ is a splitting field of $f$ and so $K = ℚ(α)$.

I would try the following line of argument. Only outlining it now (think of it as extended hints):

1. Identify the Galois group $G$ as a subgroup of the permutation of the $n$ roots. Why is $G$ transitive?
2. Let $G_1, G_2, \ldots, G_n$ be the point stabilizers. That is, if $x_i$, $i=1,2,\ldots,n$, is one of the roots, then $G_i=\{\sigma\in G\mid \sigma(x_i)=x_i\}$. Show that all the groups $G_i$ are conjugate to each other and hence equal to each other.
3. Show that the intersection $\cap_iG_i$ is trivial, and conclude that all the subgroups $G_i$ are trivial.
4. Show that $|G|=n|G_i|$ for any $i$. Enjoy the happiness that comes from having completed this task.