= Solution
The <special linear group over a finite field> $SL_n(q)$ consists of the determinant-one invertible matrices over $\mathbb F_q$. Counting an ordered <basis> gives
$$
|GL_n(q)|=\prod_{i=0}^{n-1}(q^n-q^i)
=q^{n(n-1)/2}\prod_{j=1}^n(q^j-1).
$$
The <determinant> map is onto $\mathbb F_q^*$, so
$$
\boxed{|SL_n(q)|=q^{n(n-1)/2}\prod_{j=2}^n(q^j-1).}
$$
A <transvection> has the form $I+vf$, where $v\ne0$, the <linear functional> $f$ is nonzero and $f(v)=0$. It has <determinant> one, fixes the <hyperplane> $\ker f$ pointwise, and its displacement has image the line $\langle v\rangle$. In particular $E_{ij}(t)=I+te_{ij}$ for $i\ne j$ is an elementary <transvection> when $t\ne0$.
Row addition is multiplication by such a <matrix>. <Gaussian elimination> reduces a determinant-one <matrix> to diagonal form using row additions and determinant-one pivot exchanges. The latter, and all remaining diagonal factors, are themselves products of <transvections>: within two coordinates,
$$
w(t)=E_{12}(t)E_{21}(-t^{-1})E_{12}(t)=\begin{pmatrix}0&t\\-t^{-1}&0\end{pmatrix},\qquad
w(t)w(-1)=\operatorname{diag}(t,t^{-1}).
$$
A diagonal <matrix> with product of entries one is a product of these two-coordinate diagonal matrices. Thus \b[<transvections> generate $SL_n(q)$]. This generation statement actually holds for every $n\ge2$ and every <finite field>; the exclusions in the question are needed for the subsequent simplicity assertion.
Commuting with all elementary matrices forces the <group center> to consist of scalar matrices $\lambda I$ with $\lambda^n=1$, so its order is $d=\gcd(n,q-1)$. The projective quotient acts faithfully on the one-dimensional subspaces. Its action is two-transitive: send an ordered pair of independent representative vectors to any other pair and adjust the <determinant> by rescaling a representative without changing its line. Therefore it is primitive.
The <Iwasawa simplicity lemma> says that in a faithful primitive action, if a <point stabilizer> has an abelian <normal subgroup> whose conjugates generate the <group>, every nontrivial <normal subgroup> contains the derived <subgroup>. In particular a nontrivial <perfect group> satisfying these conditions is simple. No proof of the lemma is required here.
For the point $\ell=\langle v\rangle$, the <subgroup>
$$
U_\ell=\{I+vf:f(v)=0\}
$$
is abelian, is normal in the line stabilizer, and its conjugates contain all <transvections>. Its image in the projective quotient has the same properties. Perfectness follows explicitly from elementary <commutators>. If $n\ge3$, choose a third index $r$ to obtain
$$
[E_{ir}(a),E_{rj}(b)]=E_{ij}(ab).
$$
If $n=2,q>3$, choose $t$ with $t^2\ne1$ and set $h(t)=\operatorname{diag}(t,t^{-1})$. Then
$$
[h(t),E_{12}(c)]=E_{12}((t^2-1)c).
$$
Every upper <transvection> is a <commutator>, and <conjugation> gives the lower ones. The <group> and its projective quotient are perfect. <Iwasawa's simplicity lemma> now proves
$$
\boxed{PSL_n(q)\text{ is simple for }n\ge3\text{ or }n=2,\ q>3.}
$$
The order of $PSL_2(4)$ is $4(4^2-1)=60$. Its faithful projective-line action has degree five, so its image is an index-two <subgroup> of $S_5$, necessarily $A_5$. Thus $PSL_2(4)\cong A_5$.
For $PSL_2(5)$ the order is $5(5^2-1)/2=60$. We construct a faithful degree-five action. In $SL_2(5)$ take
$$
i=\begin{pmatrix}2&0\\0&3\end{pmatrix},\qquad
j=\begin{pmatrix}0&1\\4&0\end{pmatrix}.
$$
They satisfy $i^2=j^2=-I$ and $ij=-ji$. Their quaternion <subgroup> projects to a Sylow 2-subgroup $V\cong C_2^2$. Its projective <centralizer> is $V$: a lift of a centralizing element conjugates each of $i,j$ to itself or its negative. In the <matrix> <basis> $I,i,j,ij$, those four sign choices each leave a one-dimensional space, and <determinant> one leaves just its two quaternion representatives. The <matrix>
$$
r=\frac{-I+i+j+ij}{2}=\begin{pmatrix}3&4\\3&1\end{pmatrix}
$$
has order three and cyclically permutes the three nonidentity elements of $V$. The <normalizer> quotient embeds in $S_3$ and has order dividing both $6$ and $60/4=15$, so it has order exactly three. Hence $|N(V)|=12$ and there are five Sylow 2-subgroups. Their <conjugation> action is nontrivial and transitive, and simplicity makes it faithful. Its image again has order $60$ in $S_5$, proving
$$
\boxed{PSL_2(4)\cong PSL_2(5)\cong A_5.}
$$
Finally,
$$
|PSL_3(4)|=\frac{4^3(4^2-1)(4^3-1)}3=20160,qquad
|PSL_4(2)|=2^6(2^2-1)(2^3-1)(2^4-1)=20160.
$$
In <characteristic two> an involutory <matrix> has form $I+N$ with $N^2=0$. Its Jordan blocks have size at most two. A projective <involution> in $PSL_3(4)$ has a unique involutory lift to $SL_3(4)$: its square is central, and squaring is an automorphism of the scalar center of order three, allowing a unique scalar adjustment. The nonidentity Jordan form in <dimension> three is only $(2,1)$. This GL-class remains a single SL-class, since the <centralizer> contains $\operatorname{diag}(I_2,c)$ of every <determinant> $c\in\mathbb F_4^*$. Thus $PSL_3(4)$ has one <involution> class.
In $PSL_4(2)=SL_4(2)=GL_4(2)$ there are two forms, $(2,1,1)$ and $(2,2)$, so there are two <involution> classes. More concretely, the first <group>'s single class has $21\cdot5\cdot3=315$ elements, counting image line, <kernel> plane through it and nonzero map between the quotient and image. The rank-one class in <dimension> four over $\mathbb F_2$ has $15\cdot7=105$ elements; the rank-two class has $35\cdot|GL_2(2)|=210$. <Isomorphisms> preserve conjugacy classes, so
$$
\boxed{PSL_3(4)\not\cong PSL_4(2),\quad\text{despite their common order }20160.}
$$
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