Minimal generation of simple groups and Ore's conjecture The well known Ore's conjecture (now established) states that every element of a finite non-abelian simple group $G$ is a commutator of a pair of elements. Also we know that $G$ is $2$-generated.
I am trying to find out what is known about: given any $1 \neq x \in G$, can it be a commutator of two generating elements, i.e., $x = [a,b]$ so that $G$ is generated by $a, b$ as well.
If the answer is negative, are there known restrictions on the conjugacy class of $x$ for which this happens?
The question is motivated from the action of $G$ on Riemann surfaces that yield orbit genus $1$ corresponding to minimal signatures for the group. 
 A: This question was answered on another forum, so I will just repeat the answer from there.
It is true for 'most' finite simple groups, but there are lots of exceptions, including ${\rm PSL}(2,2^n)$ for all $n$, ${\rm PSL}(3,3)$, ${\rm PSU}(3,3)$, $A_8$, ${\rm PSp}(4,3)$, and $M_{11}$. It is not true in general in ${\rm SL}(n,q)$ and ${\rm Sp}(2n,q)$, so they are also exceptions whenever they have trivial centre.
In particular, $A_5$ is an exception. If $a,b \in A_5$ with $[a,b]$ of order $2$, then $\langle a,b \rangle \cong A_4$.
A: Here is a character-theoretic argument which shows that for each $n > 1$, whenever an involution $t \in {\rm SL}(2,2^{n})= G$ has $t = [a,b]$, then $a,b \in B = N_{G}(S)$, where $S$ is the unique Sylow $2$-subgroup of $G$ containing $t$.
In general, when $G$ is a finite group and $x \in G$, then the number of ordered pairs $(a,b) \in G \times G$ with $x = [a,b]$ is expressible as $\sum_{ \chi \in {\rm Irr}(G)} \frac{|G|\chi(x)}{\chi(1)},$ where ${\rm Irr}(G)$ is the set of complex irreducible characters of $G$. This formula was probably known to W. Burnside (in fact, the fact that $x$ is a commutator if and only if the sum is positive appears in Burnside's book, and was later important in the proof of the Ore conjecture).
Letting $T,S, B, G$ be as above, we note that $B$ is a Frobenius group of order $2^{n}(2^{n}-1)$ , and has $2^{n}-1$ irreducible characters of degree $1$ (each with $t$ in their kernel), and one irreducible character of degree $2^{n}-1$ taking value $-1$ at $t$. Hence the number of order pairs $(a,b) \in B \times B$ with $t = [a,b] $ is 
$(2^{n}(2^{n}-1) [ (2^{n}-1) - \frac{1}{2^{n}-1}] = 2^{3n}-2^{2n+1}$.
On the other hand, $G$ has one irreducible character of degree $1$, the trivial character, one irreducible character of degree $2^{n}$ (which vanishes at $t$), $2^{n-1}$ irreducible characters of degree $2^{n}-1$, all taking value $-1$ at $t$, and $2^{n-1}-1$ irreducible characters of degree $2^{n}+1$, all taking value $1$ at $t$.
Hence the number of ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ is 
$2^{n}(2^{2n}-1) [ 1 - \frac{2^{n-1}}{(2^{n}-1)} + \frac{2^{n-1}-1}{2^{n}+1}],$ which also turns out to be $2^{3n} - 2^{2n+1}$. 
Hence all ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ actually lie in $B \times B$, so it is not possible to express $t = [a,b]$ where $\langle a,b \rangle = G.$
