Showing $\mathrm{PSL}(2,q)$ is perfect Let $G = \mathrm{PSL}(2,q)$ be given as permutation group on $\Omega = \mathbb{F}_q \cup \{\infty\}$, where $q = p^n$ is a prime power. $G$ is generated by translations $\alpha_t \colon x \mapsto x+t$ ($t \in \mathbb{F}_q$), multiplications with squares $\beta_k \colon x \mapsto k^2 x$ ($k \in \mathbb{F}_q^\times$), and by $\gamma \colon x \mapsto -x^{-1}$. I am looking for a direct proof that $\mathrm{PSL}(2,q)$ is perfect for $q > 3$, i.e. G' = G.
My attempt: If $q>3$ then any translation $\alpha_t$ is a commutator, so $G'$ contains the subgroup $P \leq G$ of all translations (which is also a $p$-Sylow subgroup of $G$). Hence $U := \langle P^g : g \in G \rangle \leq G'$. It suffices to show $U=G$. Simple calculations show that $U$ has at least $q^2$ elements, and that $U$ is 2-transitive on $\Omega$. In particular, $G = U \langle \beta_k \rangle$, where $k \in \mathbb{F}_q^\times$ is primitive, so it suffices to show $\beta_k \in U$.
I would be happy about any hint how to proceed. Different attempts are appreciated as well.
 A: I know you said that you wanted to avoid matrices, but frankly that seems silly, especially given that the definition of ${\rm PSL}(2,q)$ involves matrices. Why tie your hands behind your back?
It is easier to show that $G = {\rm SL}(2,q)$ is generated by its Sylow $p$-subgroups than to do the same directly for ${\rm PSL}(2,q)$. So for $g \in G$ we want to show that $G$ is a product of elements of order $p$.  
Now $G$ acts on a $2$-dimensional vectors space $V$ over ${\mathbb F}_q$. Let $v \in V$. Since not every $p$-element fixes the subspace generated by $v$, by multiplying $g$ by a $p$-element if necessary, we may assume that $v$ and $u=gv$ are linearly independent.
Now there is s $p$-element $h$ with $h(v-u) = v-u$ and $hu = u + (v-u) = v$, and so $hgv = v$ and, since $hg$ has determinant $1$, it must be $p$-element, and we are done.
Note that one part of the more general proof that ${\rm SL}(n,q)$ is perfect is to show that it is generated by transvections, and that is what we have done here.
