  If $\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 $\Gamma $ is hereditary then $\Lambda $ has a Brauer
tree (under modest additional assumptions). Otherwise $\Delta := \Id(J(\Gamma))
= (J(\Gamma)^2)^{#} $. If $\Lambda = \Z_p G$ for a $p$-group $G\neq 1$, then
$\Gamma $ is hereditary iff $G\cong C_p$ and otherwise $[\Delta : \Lambda ] =
p^2 | G/(G'G^p)| $. For Abelian groups $G$, the length of the radical idealizer
chain of $\Z_pG$ 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 $G$.
