Recall that $|G| = [L : K] = nef$, see Lemma \ref{lemma-galois-conclusion}. Since $G$ acts transitively on the set $\{\mathfrak m_1, \ldots, \mathfrak m_n\}$ of maximal ideals of $B$ (Lemma \ref{lemma-galois}) and since $D$ is the stabilizer of an element we see that $|D| = ef$. By Lemma \ref{lemma-galois-galois} we have $$ ef = |D| = |I| \cdot |\text{Aut}(\kappa(\mathfrak m)/\kappa)| $$ where $\kappa$ is the residue field of $A$. As $\kappa(\mathfrak m)$ is normal over $\kappa$ the order of $\text{Aut}(\kappa(\mathfrak m)/\kappa)$ differs from $f$ by a power of $p$ (see Fields, Lemma \ref{fields-lemma-normal-and-automorphisms} and discussion following Fields, Definition \ref{fields-definition-insep-degree}). Hence the prime to $p$ part of $|I|$ is equal to the prime to $p$ part of $e$. \medskip\noindent Set $C = B_\mathfrak m$. Then $I$ acts on $C$ over $A$ and trivially on the residue field of $C$. Let $\pi_A \in A$ and $\pi_C \in C$ be uniformizers. Write $\pi_A = u \pi_C^e$ for some unit $u$ in $C$. For $\sigma \in I$ write $\sigma(\pi_C) = \theta_\sigma \pi_C$ for some unit $\theta_\sigma$ in $C$. Then we have $$ \pi_A = \sigma(\pi_A) = \sigma(u) (\theta_\sigma \pi_C)^e = \sigma(u) \theta_\sigma^e \pi_C^e = \frac{\sigma(u)}{u} \theta_\sigma^e \pi_A $$ Since $\sigma(u) \equiv u \bmod \mathfrak m_C$ as $\sigma \in I$ we see that the image $\overline{\theta}_\sigma$ of $\theta_\sigma$ in $\kappa_C = \kappa(\mathfrak m)$ is an $e$th root of unity. We obtain a map \begin{equation} \label{equation-inertia-character} \theta : I \longrightarrow \mu_e(\kappa(\mathfrak m)),\quad \sigma \mapsto \overline{\theta}_\sigma \end{equation} We claim that $\theta$ is a homomorphism of groups and independent of the choice of uniformizer $\pi_C$. Namely, if $\tau$ is a second element of $I$, then $\tau(\sigma(\pi_C)) = \tau(\theta_\sigma \pi_C) = \tau(\theta_\sigma) \theta_\tau \pi_C$, hence $\theta_{\tau \sigma} = \tau(\theta_\sigma) \theta_\tau$ and since $\tau \in I$ we conclude that $\overline{\theta}_{\tau \sigma} = \overline{\theta}_\sigma \overline{\theta}_\tau$. If $\pi'_C$ is a second uniformizer, then we see that $\pi'_C = w \pi_C$ for some unit $w$ of $C$ and $\sigma(\pi'_C) = w^{-1}\sigma(w)\theta_\sigma \pi'_C$, hence $\theta'_\sigma = w^{-1}\sigma(w)\theta_\sigma$, hence $\theta'_\sigma$ and $\theta_\sigma$ map to the same element of the residue field as before. \medskip\noindent Since $\kappa(\mathfrak m)$ has characteristic $p$, the group $\mu_e(\kappa(\mathfrak m))$ is cyclic of order at most the prime to $p$ part of $e$ (see Fields, Section \ref{fields-section-roots-of-1}). \medskip\noindent Let $P = \Ker(\theta)$. The elements of $P$ are exactly the elements of $D$ acting trivially on $C/\pi_C^2C \cong B/\mathfrak m^2$. Thus (a) is true. This implies (b) as $P$ is the kernel of the map $D \to \text{Aut}(B/\mathfrak m^2)$. If we can prove (c), then parts (d) and (e) will follow as $I_t$ will be isomorphic to $\mu_e(\kappa(\mathfrak m))$ as the arguments above show that $|I_t| \geq |\mu_e(\kappa(\mathfrak m))|$. \medskip\noindent Thus it suffices to prove that the kernel $P$ of $\theta$ is a $p$-group. Let $\sigma$ be a nontrivial element of the kernel. Then $\sigma - \text{id}$ sends $\mathfrak m_C^i$ into $\mathfrak m_C^{i + 1}$ for all $i$. Let $m$ be the order of $\sigma$. Pick $c \in C$ such that $\sigma(c) \not = c$. Then $\sigma(c) - c \in \mathfrak m_C^i$, $\sigma(c) - c \not \in \mathfrak m_C^{i + 1}$ for some $i$ and we have \begin{align*} 0 & = \sigma^m(c) - c \\ & = \sigma^m(c) - \sigma^{m - 1}(c) + \ldots + \sigma(c) - c \\ & = \sum\nolimits_{j = 0, \ldots, m - 1} \sigma^j(\sigma(c) - c) \\ & \equiv m(\sigma(c) - c) \bmod \mathfrak m_C^{i + 1} \end{align*} It follows that $p | m$ (or $m = 0$ if $p = 1$). Thus every element of the kernel of $\theta$ has order divisible by $p$, i.e., $\Ker(\theta)$ is a $p$-group.