= Solution
Let $V=F^n$ and let $G=PSL(n,F)$ act on its one-dimensional subspaces. First identify the <kernel of a group homomorphism> in $SL(n,F)$. A <matrix> fixing every coordinate line is diagonal, and fixing every line $\langle e_i+e_j\rangle$ forces all diagonal entries equal. Thus the projective <kernel of a group homomorphism> is $\{\lambda I:\lambda^n=1\}$. These scalars are central. Conversely a central <matrix> commutes with all $E_{ij}(1)$: comparing entries in $Me_{ij}=e_{ij}M$ shows all off-diagonal entries of $M$ vanish and all diagonal entries agree. Therefore the <kernel of a group homomorphism> is exactly $Z(SL(n,F))$. This proves <scalar kernel of the projective linear action> and faithfulness of the induced $G$-action.
The action is doubly transitive. Given two ordered pairs of distinct lines, choose two independent representatives of each pair and extend them to <bases>. A <linear map> taking the first <basis> to the second sends the desired lines correctly. If its <determinant> is $d\ne1$, postcompose with a diagonal map in the target <basis> that scales its first <basis> <vector> by $d^{-1}$ and fixes all the others. It preserves the two target lines and corrects the <determinant>. Thus an element of $SL(n,F)$ sends either ordered pair to the other. Double transitivity implies primitivity: a block containing two distinct points must contain every point by the transitive point-stabilizer action.
For $\alpha=\langle e_1\rangle$, consider
$$
U=\{I+e_1\varphi:\varphi\in V^*,\ \varphi(e_1)=0\}.
$$
Products add the functionals, so $U$ is abelian. If $g$ fixes $\alpha$, write $ge_1=ae_1$. Then $g(I+e_1\varphi)g^{-1}=I+e_1(a\varphi g^{-1})$, and the new functional still vanishes on $e_1$. Thus the image $A$ of $U$ in $G$ is normal in the <point stabilizer>. Its conjugates contain all elementary <transvections>: changing the center line $\langle e_1\rangle$ to $\langle e_i\rangle$ gives every map $I+e_i\psi$ with $\psi(e_i)=0$, including $E_{ij}(t)$. Part (b) then shows that those conjugates generate $G$.
It remains to prove perfectness, not assume it. Use $[x,y]=x^{-1}y^{-1}xy$. For $n\geq3$, distinct $i,j,k$ give
$$
[E_{ik}(t),E_{kj}(1)]=E_{ij}(t).
$$
Every generator is therefore a <group commutator>, so $SL(n,F)'=SL(n,F)$. For $n=2$ and $|F|>3$, choose $a\in F^\times$ with $a^2\ne1$, possible because a quadratic polynomial has at most two roots. With $D=\operatorname{diag}(a,a^{-1})$,
$$
[D,E_{12}(t)]=E_{12}((1-a^{-2})t),\qquad
[D,E_{21}(t)]=E_{21}((1-a^2)t).
$$
Both coefficients are nonzero, so every upper and lower elementary <transvection> is again a <group commutator>. Part (b) proves perfectness in this case as well. Quotients of perfect <groups> are perfect, hence $G'=G$ in exactly the stated range.
Apply part (a) to the faithful primitive $G$-action and its abelian normal point-stabilizer <subgroup> $A$. Every nontrivial <normal subgroup> contains $G'=G$, so it is the whole <group>. A <transvection> is not scalar, showing $G\ne1$. We have proved
$$
\boxed{PSL(n,F)\text{ is simple for }n\geq3,\text{ or }n=2\text{ with }|F|>3.}
$$
The proof works for infinite <fields> too and uses no finite-order count. The excluded two-dimensional <fields> give the familiar soluble exceptions $PSL(2,2)\cong S_3$ and $PSL(2,3)\cong A_4$.
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