Let $L/K$ be a finite Galois extension where $K= \operatorname{Frac}(O)$ is complete and $O$ is a Dedekind domain. Let $\Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$. Then let $\Delta_i=\{\delta\in\Delta\vert\forall a\in O_L,\delta(a)-a\in Q^i\}$.
$\Delta_1$ is unique $p$-Sylow subgroup of $\Delta_0$. It is clear that $\Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $\Delta_1$ is unique $p$-Sylow of $\Delta_0$, then $\Delta_1$ is normal by all $p$-Sylow being conjugates.
$\Delta_0$ is normal due to exact sequence $1\to\Delta_0\to\Delta\to \operatorname{Gal}(k_L/k)\to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.
$\textbf{Q:}$ The book says it is follows easily $\Delta_1\leq\Delta$ is normal. Why $\Delta_1$ is normal here? Normality is non-transitive. I knew $\Delta_1\leq\Delta_0$ normal and $\Delta_0\leq\Delta$ normal. $ef=[L:K]$. I knew $|\Delta_0|=e$. However there is no reason to expect $p\nmid f$.
Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.