September 21st
Today I learned some applications of the idea that prime-splitting can tell us useful information about Galois groups. For example, suppose we have a normal extension $L/K$ with primes $\mf q/\mf p$ so that $\mf p$ splits completely in every intermediate nontrivial subfield between $L$ and $K$ but not in $L.$
Then we claim that $D=D(\mf q/\mf p)$ is the unique smallest nontrivial subgroup of $\op{Gal}(L/K).$ For this we use the argument suggested in $\textit{Number Fields}.$ We are given that $e(\mf q/\mf p)f(\mf q/\mf p) \gt 1$ because $\mf p$ does not split completely in $L,$ so we get that $[L:L_D] \gt 1.$ So it follows $D$ is actually a nontrivial subgroup. Further, for any other subgroup $H\ne\langle e\rangle$ of $G,$ there is an intermediate field $L_H\le L$ with $\mf q_H$ between $\mf p$ and $\mf q.$ So it follows $e(\mf q_H/\mf p)f(\mf q_H/\mf p)=1,$ which implies\[[L:L_EL_H]=e(\mf q/\mf q_H)f(\mf q/\mf q_H)=e(\mf q/\mf p)f(\mf q/\mf p)=[L:L_E].\]It follows that $L_H\subseteq L_E,$ so $E\subseteq H.$
It might not appear a very strong condition that we have a unique smallest nontrivial subgroup, but there are quite a few things that we get out of this. For example, we get that $D$ must be a subset and therefore equal to all of its conjugate subgroups, so $D$ is a normal subgroup. Further, considering any prime $p$ dividing into the order of $\op{Gal}(L/K),$ we know that the $D$ is a subgroup of any subgroup of order $p$ (guaranteed by Cauchy's Theorem), so in fact the Galois group is of prime-power order!
Sure, the condition is esoteric, but it ought be not terribly odd to think that perfect knowledge of prime-splitting should give perfect knowledge of the field extension; after all, the above discussion centers around information about a single prime $\mf p\subseteq K.$