Recognizing when a tower of Galois extensions gives a Galois extension

This isn't an answer, nor is it very general, just an illustration with an example. Let $K$ be a number field, $E/K$ finite Galois, and $L/E$ the Hilbert class field of $E$ (the maximal unramified abelian extension of $E$, which I'll take to be inside some algebraic closure $\overline{K}$ of $K$). Then $L/E$ is Galois by definition, and using its characterizing property, it can be proved that $L/K$ is Galois. Namely, let $\sigma:L\rightarrow\overline{K}$ be a $K$-monomorphism. Then $\sigma(E)\subseteq E$ because $E/K$ is Galois, so $\sigma(E)=E$. Thus $E\subseteq \sigma(L)$, and $\sigma$ sets up an isomorphism between the extensions $L/E$ and $\sigma(L)/E$. I don't mean that $L$ and $\sigma(L)$ are $E$-isomorphic, because, while $\sigma(E)=E$, it needn't be the case that $\sigma$ fixes $E$ pointwise. But, for example, $\sigma(L)/E$ must be Galois, and we have an isomorphism $\mathrm{Gal}(L/E)\cong\mathrm{Gal}(\sigma(E)/E)$ induced by $\sigma$. One can also verify that $\sigma(L)/E$ is unramified because $L/E$ is. So $\sigma(L)/E$ is abelian and unramified. This means $L\sigma(L)$ (compositum inside $\overline{K}$) is abelian and unramified over $E$. By maximality, $L\sigma(L)=L$, which implies that $\sigma(L)=L$. So we can conclude that $L/K$ is Galois.

I don't really know if there is a way to make this formal. Similar arguments can be used to prove that various class fields of $E$ are Galois over $K$. I guess the informal idea is that $L/E$ should be maximal (inside $\overline{K}$) with respect to some properties which are preserved by $K$-embeddings of $L$ into $\overline{K}$ (which necessarily send $E$ onto itself) and which are preserved under compositum. I admit this is vague, but I'm not sure how to make it more precise.