# 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.

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There are certain diffeomorphism groups that are perfect and have non-trivial quasi-morphisms, which in fact implies non-zero stable commutator length: arxiv.org/abs/1105.4443 –  Ian Agol May 1 '12 at 22:29

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.

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Perfect Roberto, thanks a lot, this paper answers my question. –  Portland May 2 '12 at 1:22

For finite groups, there are no examples this was the Ore conjecture How did "Ore's Conjecture" become a conjecture?.

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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.