\operatorname{disc} \mathcal{O}_K[G],$$ and writing $\underline{m} = \pi_K^A \mathcal{O}_K$ we obtain $$\label{eq:IndexIdealNumber}
[\mathfrak{M} : \mathcal{O}_K[G]] = N(\underline{m})= p^{f_KA}.$$ We now show that $\operatorname{disc} \mathcal{O}_K[G]=|G|^{|G|} \mathcal{O}_K$. We claim that the trace of an element $g \in G \subset \mathcal{O}_K[G]$ is 0 if $g \neq \operatorname{Id}$ and is $|G|$ for $g=\operatorname{Id}$. To see this, notice that an $\mathcal{O}_K$-basis of $\mathcal{O}_K[G]$ is given by $\{g \in G\}$, and in this basis the left multiplication action by $g$ is represented by a permutation matrix, whose trace equals the number of fixed points of the map $h \mapsto gh$. The group axioms imply that either $h \mapsto gh$ has no fixed points ($g \neq \operatorname{Id}$), or every element of $G$ is fixed ($g=\operatorname{Id}$). This proves the claim. In particular, the Gram matrix of the bilinear form $(x,y) \mapsto \operatorname{tr}(x,y)$ in this basis is $$\left( \operatorname{tr}(gh) \right)_{g, h \in G} = |G| \left(  \delta_{gh,\operatorname{id}} \right)_{g, h \in G},$$ whose determinant is clearly $\pm |G|^{|G|}$, because $\left(  \delta_{gh,\operatorname{id}} \right)_{g, h \in G}$ is a permutation matrix. From Equation we obtain $$\underline{m}^2 = \frac{|G|^{|G|} \mathcal{O}_K}{\operatorname{disc} \mathfrak{M}} \bigm\vert |G|^{|G|} \mathcal{O}_K,$$ so (writing again $\underline{m}=\pi_K^{A}\mathcal{O}_K$) we have $A {\leqslant}\frac12|G| e_K v_p(|G|)$. Via Equation , this implies $$\label{eq:IndexMaximalOrderBound}
v_p\left(\#\frac{\mathfrak{M}}{\mathcal{O}_K[G]}\right) {\leqslant}\frac{1}{2} [K:\mathbb{Q}_p] \cdot 