I have just added to my notes on representation theory the very cute formula of Frobenius that gives, in terms of irreducible characters, the number $latex N(g)$ of representations of a given element $latex g$ as a commutator $latex g=[x,y]=xyx^{-1}y^{-1}$ in a finite group $latex G$:
$latex N(g)=|G|\sum_{\chi}\frac{\chi(g)}{\chi(1)},$
where $latex \chi$ runs over the irreducible (complex) characters of $latex G$ (this is Proposition 4.4.3 on page 118 of the last version of the notes).
I wanted to mention some applications, and had a vague memory that this was used to show that most or all elements in various simple groups are actual commutators. By searching around a bit, I found out easily that, indeed, there was a conjecture of Ore from 1951 to the effect that the set of commutators is equal to $latex G$ for any non-abelian finite simple group $latex G$, and that (after various earlier works) this has recently been proved by Liebeck, O’Brien, Shalev and Tiep.
I mentioned this of course, but then I also wanted to give some example of non-commutator, and decided to look for this using Magma (the fact that I am recovering from a dental operation played a role in inciting me to find something distracting to do). Here’s what I found out.
First, a natural place to look for interesting examples is the class of perfect groups, of course not simple. This is also easy enough to implement since Magma has a database of perfect groups of “small” order. Either by brute force enumeration of all commutators or by implementing the Frobenius formula, I got the first case of a perfect group $latex G$, of order $latex 960$, which contains only $latex 840$ distinct commutators.
Then I wanted to know “what” this group really was. Magma gave it to me as a permutation group acting on $latex 16$ letters, with an explicit set of $latex 6$ generators, and with a list of $latex 21$ relations, which was not very enlightening. However, looking at a composition series, it emerged that $latex G$ fits in an exact sequence
$latex 1\rightarrow (\mathbf{Z}/2\mathbf{Z})^4\rightarrow G\rightarrow A_5\rightarrow 1.$
This was much better, since after a while it reminded me of one of my favorite types of groups: the Weyl groups $latex W_{g}$ of the symplectic groups $latex \mathrm{Sp}_{2g}$ (equivalently, the “generic” Galois group for the splitting field of a palindromic rational polynomial of degree $latex 2g$), which fit in an relatively similar exact sequence
$latex 1\rightarrow (\mathbf{Z}/2\mathbf{Z})^g\rightarrow W_g\rightarrow S_g\rightarrow 1.$
From there, one gets a strong suspicion that $latex G$ must be the commutator subgroup of $latex W_5$, and this was easy to check (again with Magma, though this is certainly well-known; the drop of the rank of the kernel comes from looking at the determinant in the signed-permutation $latex 5$-dimensional representation, and the drop from $latex S_5$ to $latex A_5$ is of course from the signature.)
This identification is quite nice, obviously. In particular, it’s now possible to identify concretely which elements of $latex G$ are not commutators. It turns out that a single conjugacy class, of order $latex 120$, is the full set of missing elements. As a signed permutation matrix, it is the conjugacy class of
$latex g=\begin{pmatrix} 0& -1 & 0 & 0 & 0\\ 1& 0 & 0 & 0 & 0\\ 0& 0 & 0 & 1 & 0\\ 0& 0 & 1 & 0 & 0\\ 0& 0 & 0 & 0 & -1\end{pmatrix},$
and the reason it is not a commutator is that Magma tells us that all commutators in $latex G$ have trace in $latex \{-3,-2,0,1,2,5\}$ (always in the signed-permutation representation). Thus the trace $latex -1$ doesn’t fit…
At least, this is the numerical reason. I feel I should be able to give a theoretical explanation of this, but I haven’t succeeded for the moment. Part of the puzzlement is that this behavior seems to be special to $latex W_5$, the Weyl group of the root system $latex C_5$. Indeed, for $latex g\in\{2,3,4\}$, the corresponding derived subgroup is not perfect, so the question does not arise (at least in the same way). And when $latex g\geq 6$, the derived subgroup $latex G_g$ of $latex W_g$ is indeed perfect, but — experimentally! — it seems that all elements of $latex G_g$ are then commutators.
I haven’t found references to a study of this Ore-type question for those groups, so I don’t know if these “experimental” facts are in fact known to be true. Another question seems natural: does this special fact have any observable consequence, for instance in Galois theory? I don’t see how, but readers might have better insights…
(P.S. I presume that GAP or Sage would be equally capable of making the computations described here; I used Magma mostly because I know its language better.
P.P.S And the computer also tells us that even for the group $latex G$ above, all elements are the product of at most two commutators, which a commenter points out is also a simple consequence of the fact that there are more than $latex 480$ commutators….
P.P.P.S To expand one of my own comments: the element $latex g$ above is a commutator in the group $latex W_5$ itself. For instance $latex g=[x,y]$ with
$latex x=\begin{pmatrix} 0& 0 & 0 & 0 & -1\\ 0& 1 & 0 & 0 & 0\\ 1& 0 & 0 & 0 & 0\\ 0& 0 & 1 & 0 & 0\\ 0& 0 & 0 & 1 & 0\end{pmatrix},$
and
$latex y=\begin{pmatrix} 1& 0 & 0 & 0 & 0\\ 0& 0 & 0 & 0 & -1\\ 0& 1 & 0 & 0 & 0\\ 0& 0 & 1 & 0 & 0\\ 0& 0 & 0 & -1 & 0\end{pmatrix},$
where $latex y\notin G$.)
