|G| \cdot v_p(|G|).$$ We summarise these results in the following Proposition.

\[prop:MaximalOrder\] Let $\mathfrak{M}$ be a maximal order of $K[G]$. We have $[\mathfrak{M} : \mathcal{O}_K[G]]_{\mathcal{O}_K}=\pi_K^{A}\mathcal{O}_K$ with $A=\frac{1}{2}|G|e_K v_p(|G|) - v_K(\operatorname{disc} \mathfrak{M} ) {\leqslant}\frac{1}{2}|G|e_K v_p(|G|)$.

Proof of Theorem \[thm:IntroGeneralBound\]  {#sect:GeneralBound}
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In this section we prove a bound on $m(L/K)$ which holds in complete generality. Let $\mathfrak{B}$ be an $\mathcal{O}_K$-order in $K[G]$ containing $\mathcal{O}_K[G]$. We define $\mathfrak{B}\mathcal{O}_L$ to be the $\mathfrak{B}$-lattice generated by $\mathcal{O}_L$, that is, $$\mathfrak{B} \mathcal{O}_L=\left\{\sum_i b_ix_i\right\}_{b_i\in \mathfrak{B},x_i\in\mathcal{O}_L}\subseteq L,$$ and suppose that $\mathfrak{B}  \mathcal{O}_L$ is a free $\mathfrak{B}$-module of rank $1$. We note that every maximal order $\mathfrak{B}$ has this property. Indeed, let $\mathfrak{B}$ be a maximal order of the separable $K$-algebra $K[G]$; by the normal basis theorem $K\mathfrak{B}\mathcal{O}_L=L$ is isomorphic to $K\mathfrak{B}=K[G]$ as a left $K[G]$-module, so $\mathfrak{B}\mathcal{O}_L$ is isomorphic to $\mathfrak{B}$ as a left $\mathfrak{B}$-module by [@MR1972204 Theorem 18.10].

The set $\mathfrak{B} \mathcal{O}_L$ is a fractional ideal in $L$.

Indeed, given $\sum_{g \in G} k_g g$ in $\mathfrak{B}$ (where $k_g \in K$) and $x,y \in \mathcal{O}_L$ we have $$\left( \sum_{g \in G} k_g g x \right)   y = \sum_{g \in G} k_g g \left( x g^{-1}(y) \right) \in \mathfrak{B}\mathcal{O}_L.$$

In the light of this lemma we can 