Efficient method to find the center of the $SL(2, 3)$

Note that if $G$ is a group with an abelian subgroup $H$ and $K$ is a subgroup of $H$, then $H \subseteq C_G(K)$, because then we have $hkh^{-1}=k$ for all $k \in K, h \in H$. In your example, we have that if $P$ is the 3-Sylow subgroup generated by $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$ $$H \subseteq C_G(P) \subseteq N_G(P)$$ Also, there are four 3-Sylow subgroups, so by the Sylow theorems and orbit-stabilizer $N_G(P)$ has order $\frac{24}{4}=6$. Thus we actually have $H=N_G(P)$. Note that we also have $Z(G) \subseteq C_G(P) = H$, so the order of $Z(G)$ is either $2$ or $6$. But if $Z(G)=H$, then $P \subset H$ would be a normal subgroup (because any subgroup of the center is normal), contradicting the fact that there are four 3-Sylows.