Projective special linear group over the field with five elements ($PS{L}_2(5)\cong A_5$)

The special linear group $S{L}_2(5)$ has $24$ nonzero first columns and five second columns giving determinant one, hence order $120$; its centre is $\{I,-I\}$. In the quotient by this centre, the nonidentity conjugacy classes have sizes $15,20,12,12$, corresponding to projective element orders $2,3,5,5$. These arise from trace zero, paired traces $\pm 1$, and paired unipotent traces $\pm 2$. No proper union of these classes with the identity has size dividing $60$, so the quotient is simple. The simple groups with order a power of two times fifteen result identifies it with $A_5$.

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 $A_5$.