I am migrating this question from math stackexchange...


I have a semidirect product and I have shown that it is perfect. However, I would like to know whether every element is a simple commutator (rather than a product of commutators). This is closely related to Ore's Conjecture (which is actually a theorem now, cf. Liebeck et. al), but I am not aware of any general results in this direction, so if someone could provide one, I would be grateful.


In any event, I will describe the specific setup I have below, and if someone would like to offer specific insights, I would, likewise, be appreciative.


Start with any finite field $\mathbb{F}$. Let $m\geq 1$ be an integer and consider $Sp\_{2m}(\mathbb{F})$, the group of symplectic matrices (over $\mathbb{F}$) with the natural (right) action on the vector space $\mathbb{F}^{2m}$.


Every element $v\in \mathbb{F}^{2m}$ can be realized as $v=w^{A}-w$ for an appropriate choice of $A\in Sp\_{2m}(\mathbb{F})$ and $w\in \mathbb{F}^{2m}$. The element $w^A-w$ is a commutator in the semidirect product. Moreover, if $|\mathbb{F}|\geq 3$ and $m\geq 2$, then every element of $Sp\_{2m}(\mathbb{F})$ is a commutator (in the symplectic group).


Is this enough to show that elements of the semidirect product are commutators? If not, what would be a sufficient condition?



When are elements of a (perfect) semidirect product simple commutators?
Your general question seems too general. Here is a partial answer to your specific question.


Let $V = \mathbb{F}^{2m}$ and $G = \mathrm{Sp}\_{2m}(\mathbb{F})$. Consider some $vg \in VG$. We know that $g = [x,y]$ for some $x, y \in G$. *Assume $y$ can be chosen so that $y-1$ is invertible.* Now observe that
$$[wx,y] = x^{-1} w^{-1} w^y x^y = (-wx + wyx) [x,y]\qquad (w \in V).$$
Since $y-1$ is invertible, so is $yx-x$, so we can choose $w$ so that $-wx+wyx = v$ and hence $vg = [wx,y]$.


I think such a $y$ probably can be chosen for any given $g \in G$. If $|\mathbb F| \equiv 1 \pmod 4$, this follows from a result of Gow [1], which states that $G = C^2$ where $C$ is the single conjugacy class consists of elements $y$ such that $y^2 = -1$ (verifying Thompson's conjecture in this case). Such $y$ clearly do not have eigenvalue $1$, so $y-1$ is invertible as required, and since $C^2 = G$ there must be some $y \in C$ and $x \in G$ such that $g = y^{-x}y = [x,y]$.


[1] *Gow, Roderick*, [**Commutators in the symplectic group**](https://doi.org/10.1007/BF01187734), Arch. Math. 50, No. 3, 204-209 (1988). [ZBL0628.20037](https://zbmath.org/?q=an:0628.20037).


