We know that Ore's Conjecture is a theorem now. Does anyone know a counterexample to Ore's conjecture for perfect groups? I believe there should be one, but I dont have a counterexample. Thanks.



Ore's Conjecture for perfect groups
The counterexample referred to in the comment by NAME\_IN\_CAPS is the split extension of the so-called deleted permutation module for $A\_5$ over ${\mathbb F}\_2$. that is, the $4$-dimensional irreducible component of the $5$-dimensional permutation module. This group has $960$ elements, only $840$ of which are commutators.


Here is a general recipe for constructing examples of perfect groups in which, for any given $k$, there are elements of $[G,G]$ that are not product of at most $k$ commutators.


It can illustrated by using the other $4$-dimensional irreducible module $M$ for $A\_5$ over ${\mathbb F}\_2$, namely the non-absolutely irreducible module arising from the isomorphism $A\_5 \cong {\rm SL}(2,4)$.


Calculations show that the tensor product $M \otimes M$ maps onto $I \oplus I$, where $I$ is the trivial module. So, for any $n \ge 1$, the semidirect product $H\_n$ of $M^n$ by $A\_5$ has a perfect central extension $G\_n$ with centre elementary abelian of order $2^{n(n-1)/2}$, because we can get a $2^2$ for each pair of direct factors of $M^n$.


(The full Schur Multiplier of $H\_n$ is larger and has the structure $2 \times 4^{2n} \times 2^{n(n-1)/2}$.)


Now $|H\_n| = 2^{4n} \times 60$ and $|G\_n| = 2^{n(n-1)/2} \times |H\_n|$. Since a commutaor $[x,y]$ is not changed by multiplying $x,y$ by central elements, $G\_n$ has at most $|H|^2$ distinct commutators, and there are at most $|H|^{2k}$ elements of $G\_n$ that are products of at most $k$ commutators. This is less than $|G\_n|$ for large enough $n$.


Of course, we can perform similar constructions for other modules of other simple groups.


