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Use the group commutator convention $[x,y]=x^{-1}{y}^{-1}xy$. If $n\ge 3$, choose distinct $i,j,k$. Multiplying elementary matrices, using $E_{ik}{E}_{kj}=E_{ij}$ and all reverse or repeated products here equal to zero, givesEvery elementary generator is therefore in the commutator subgroup. Part iii shows these generators generate $\mathrm{SL}(V)$, proving it is a perfect group.

It remains to handle $n=2$ and $q>3$. The polynomial $s^2-1$ has at most two roots, whereas ${\mathbb{F}}_q^{\times }$ has more than two elements, so choose $s\ne 0$ with $s^2\ne 1$. For $h=\mathrm{diag}(s,s^{-1})$, direct conjugation givesBoth coefficients are nonzero. As $t$ varies, these group commutators give every upper and lower elementary transvection, which generate ${\mathrm{SL}}_2({\mathbb{F}}_q)$. Thus the perfectness of the special linear group is established in all requested cases:The two pairs in the PDF are exclusions, not a literal logical disjunction. They really are exceptions: ${\mathrm{SL}}_2({\mathbb{F}}_2)\cong S_3$ has a nontrivial sign quotient. Also ${\mathrm{SL}}_2({\mathbb{F}}_3)$ maps onto ${\mathrm{PSL}}_2({\mathbb{F}}_3)\cong A_4$, which has the nontrivial abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action has degree four and image of order 12, hence the index-two subgroup $A_4$ of $S_4$. Neither exceptional group is perfect.