contained in exactly two maximal orders with index $4$, described by the inclusion chains: $${\mathbb{Z}}\langle{i,j}\rangle 
  \subsetneq {\mathbb{Z}}\langle{i,\frac{1+i+j+k}{2}}\rangle
  \subsetneq 
  \left\{
  \begin{array}{l}
  {\displaystyle}{\mathbb{Z}}\langle{i,\frac{1+j}{2}}\rangle{\raisebox{0.4ex}{,}}\\[2.5mm]
  {\displaystyle}{\mathbb{Z}}\langle{i,\frac{1+k}{2}}\rangle\cdot
  \end{array}
  \right.$$ In particular ${\mathbb{Z}}\langle{i,(1+i+j+k)/2}\rangle$ is an Eichler order, but ${\mathbb{Z}}\langle{i,j}\rangle$ is not.

\[lem:disc-8\] Let $p \equiv 5 \bmod 8$ be a prime, and let $B = {\mathbb{Q}}\langle{i,j}\rangle$ be the quaternion algebra given by the presentation $i^2 = -2$, $j^2 = -p$, and $k = ij = -ji$, and set $R = {\mathbb{Z}}[i]$. Then $B$ is ramified only at $p$ and $\infty$, and ${\mathbb{Z}}\langle{i,j}\rangle$ is contained in exactly two maximal orders with index $8$, described by the inclusion chains: $${\mathbb{Z}}\langle{i,j}\rangle 
  \subsetneq {\mathbb{Z}}\langle{i,j,\frac{i+k}{2}}\rangle
  \subsetneq {\mathbb{Z}}\langle{i,\frac{i+k}{2},\frac{1+j+k}{2}}\rangle
  \subsetneq 
  \left\{
  \begin{array}{l}
  {\displaystyle}{\mathbb{Z}}\langle{i,\frac{1+j+k}{2},\frac{i+2j+k}{4}}\rangle{\raisebox{0.4ex}{,}}\\[2.5mm]
  {\displaystyle}{\mathbb{Z}}\langle{i,\frac{1+j+k}{2},\frac{i+2j-k}{4}}\rangle\cdot
  \end{array}
  \right.$$ In particular ${\mathbb{Z}}\langle{i,j}\rangle$ is not an Eichler order.

\[lem:disc-q\] Let $p$ and $q$ be primes, with $p \equiv 1 \bmod 4$, $q \equiv 3 \bmod 4$, and $${\left(\!\frac{-p}{q}\!\right)} = 1.$$ Let $B = {\mathbb{Q}}\langle{i,j}\rangle$ be the quaternion algebra given by the relations $i^2 = -q$, $j^2 = -p$, and $k = ij = -ji$, and set $R = {\mathbb{Z}}[(1+i)/2]$. Then $B$ is ramified only at $p$ and $\infty$, and ${\mathbb{Z}}\langle{(1+i)/2,j}\rangle = R + Rj$ is contained in exactly two maximal orders with index $q$, described by the 