 Quaternion construction of an index-five subgroup of PSL2 over F5

Over ${\mathbb{F}}_5$, putThese determinant-one matrices satisfy $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. Conjugation by $r$ cycles $i,ij,j$. Thus $⟨i,j⟩$ is a quaternion group of order eight, normalized by an element of order three. Their generated subgroup has order $24$ in $S{L}_2(5)$, hence order $12$ after quotienting by $\{\pm I\}$. Its index-five coset action, together with simplicity and perfectness, identifies $PS{L}_2(5)$ with the alternating group $A_5$.