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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.


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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: – Ian Agol May 1 '12 at 22:29
up vote 18 down vote accepted

For infinite groups, you may find examples here:

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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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.

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Professor Isaacs, do you happen to know which group equations (this question concerns the equation g=[x,y]) admit a character-theoretic criterion for the existence of solutions? For instance, g=[x1,y1][x2,y2]...[xn,yn] is possible, and g=x^2 is not (since it's the Frobenius-Schur indicator). – John Wiltshire-Gordon May 14 '12 at 21:11
John, No I don't know. That's an interesting question. – Marty Isaacs May 15 '12 at 23:40

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