# Percentage of elements that can be written as commutators

Let $G$ be a finite perfect group. Are there any results on the number of elements of $G$ which can be written as a commutator? When $G$ is finite non-abelian simple group, then every element can be written as a commutator.

-
It's clearly at most $|G/Z(G)|^2$, which can be arbitrarily small as a fraction of $|G|$. See mathoverflow.net/questions/179834 – Derek Holt Sep 4 '14 at 19:18

The proportion of elements of a finite group $G$ which are commutators is at least $\frac{1}{k(G)}$ (and the inequality is strict for non-Abelian groups) where $k(G)$ is the number of conjugacy classes of $G,$ because there are $|G|^{2}$ expressions of the form $[a,b]$ and every element of $G$ has at most $k(G)|G|$ expressions of the form $[a,b]$ (and only at the identity can that bound be achieved). Hence the number of distinct commutators in $G$ is at least $\frac{|G|^{2}}{k(G)|G|} = \frac{|G|}{k(G)}.$
On the other hand (apart from the obvious example of Abelian groups where the boound is clearly sharp), a non-Abelian extraspecial $2$-group $G$ of order $2^{2n+1}$ has only $2$ commutators, and has $k(G) = 2^{2n} + 1,$ so $\frac{|G|}{k(G)} \to 2$ as $n \to \infty,$ while we always have $\frac{|G|}{k(G)} < 2.$ Hence the proportion of commutators can get arbitrarily close to $\frac{1}{k(G)}$ for non-Abelian groups
Fulman and Guralnick proved that if $G$ is almost simple, then we have $k(G) <|G|^{0.41}$. Guralnick and I proved that we have $k(G) < (k(F)|G|)^{1/2}$ for an arbitrary finite group $G$, where $F = F(G)$ is the Fitting subgroup of $G$- I am not sure if the latter bound can be improved for perfect $G.$ – Geoff Robinson Sep 5 '14 at 7:09