Non-commutator in simple group? Hi,
For a group $G$, we say that $x\in G$ is a commutator if there exists $a,b \in G$ such that $x=a^{1}b^{-1}ab$, and we say that $x$ is a non-commutator if there is no $a,b \in G$ such that $x=a^{1}b^{-1}ab$.
Does there exist a non abelian simple group $G$ (finite or not) such that $G$ has at least one non-commutator?
I tried with $-I_n$ in $\mathrm{PSL}(n,q)$ but no luck.
Thanks.
 A: For finite groups, there are no examples  this was the Ore conjecture How did "Ore's Conjecture" become a conjecture?. 
A: For infinite groups, you may find examples here:
http://arxiv.org/abs/arXiv:0909.2294
In fact, in the reference above you may find examples of infinite simple groups with infinite commutator width. In other words, examples of simple groups $G$ such that for every $n\in\mathbb{N}$ there exists an element $g\in G$ that is not the product of less than $n$ commutators.
A: The proof of the Ore conjecture was recently completed by Liebeck, O'Brien, Shalev and Tiep. The proof depends heavily on the classification of simple groups.
If one weakens the condition that $G$ is nonabelian simple and assumes the much weaker condition that $G' = G$, then lots of examples exist where G contains noncommutators.
See, for example, my note in the MAA Monthly 84 (1977) 720-722.
Finally, I mention a character-theoretic condition that an element $x$ of $G$ is a noncommutator. It is that $\sum \chi(x)/\chi(1) = 0$, where the sum runs over all $\chi \in {\rm Irr}(G)$. This sum is always positive if $x$ is a commutator. 
