On the radical idealizer chain of symmetric orders

If $\mathrm{\Lambda }$ is an indecomposable, non maximal, symmetric order, then the idealizer of the radical \Gamma := \Id(J(\Lambda)) = J(\Lambda)^{#} is the dual of the radical. If $\mathrm{\Gamma }$ is hereditary then $\mathrm{\Lambda }$ has a Brauer tree (under modest additional assumptions). Otherwise $\Delta := \Id(J(\Gamma)) = (J(\Gamma)^2)^{#} $\mathrm{.}If$\Lambda = \Z_p G$fora$p$-group$G\neq 1$,then$\Gamma $ishereditaryiff$G\cong C_p$andotherwise$[\Delta : \Lambda ] = p^2 | G/(G’G^p)| $\mathrm{.}ForAbeliangroups$G$, the length of the radical idealizer chain of ${\mathbb{Z}}_{p}G$ is $(n-a)(p^a-p^{a-1})+p^{a-1}$, where $p^n$ is the order and $p^a$ the exponent of the Sylow $p$-subgroup of $G$.