= Quaternion construction of an index-five subgroup of PSL2 over F5
Over $\mathbb F_5$, put
$$
i=\begin{pmatrix}0&1\\4&0\end{pmatrix},\quad
j=\begin{pmatrix}2&0\\0&3\end{pmatrix},\quad
r=\begin{pmatrix}3&2\\1&1\end{pmatrix}.
$$
These determinant-one <matrices> satisfy $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. <Conjugation> by $r$ cycles $i,ij,j$. Thus $\langle i,j\rangle$ is a <quaternion group> of order eight, normalized by an element of order three. Their generated <subgroup> has order $24$ in $SL_2(5)$, hence order $12$ after quotienting by $\{\pm I\}$. Its index-five <coset> action, together with simplicity and perfectness, identifies $PSL_2(5)$ with the <alternating group> $A_5$.
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