= Solution
For the <general linear group> over $\mathbb F_{31}$, choose a nonzero first column in $31^2-1$ ways and a second column outside its span in $31^2-31$ ways. Thus
$$
|\operatorname{GL}_2(\mathbb F_{31})|=(31^2-1)(31^2-31)=2^7\,3^2\,5^2\,31.
$$
Since $5^3=125\equiv1\pmod{31}$ and $5\ne1$, the residue $5$ has order three. The nine <diagonal matrices> $\operatorname{diag}(5^i,5^j)$, with $0\le i,j<3$, form a subgroup isomorphic to $C_3\times C_3$. It is an <abelian group> and a <Sylow subgroup> for $p=3$. Every other <Sylow subgroup> for $p=3$ is conjugate to it by the <Sylow theorems>, hence is also <Abelian>.
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