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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?

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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, Arch. Math. 50, No. 3, 204-209 (1988). ZBL0628.20037.

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    $\begingroup$ Some mutterings: If $y$ itself doesn't work then we can replace $x$ and $y$ by $-x$ and $-y$. That will work unless both $x$ and $y$ have both $\pm1$ as eigenvalues. For example, for $m = 2$ in characteristic $\ne 2$, it seems like we might just be able to conclude by hand. In general, the only problem is if $x C_G(y)$ and $y C_G(x)$ consist entirely of elements with fixed points. It seems like this is probably hardest to rule out when both $x$ and $y$ are regular unipotent. $\endgroup$
    – LSpice
    Commented Mar 29, 2023 at 14:29
  • $\begingroup$ How do you conclude that $[wx,y]=(-wx+wyx)[x,y]$? $\endgroup$
    – Makenzie
    Commented Apr 4, 2023 at 17:28
  • $\begingroup$ @Makenzie The notation can be confusing. Particularly $vg$ can be ambiguous because $V$ is both a right $G$-module and a subgroup of $VG$. Moreover we tend to use additive notation in $V$ but multiplicative notation in $VG$. It might clearer if I wrote $[wx,y] = x^{-1} w^{-1} y^{-1} w x y = w^{-x} w^{yx} [x,y] = (-w^x + w^{yx}) [x,y]$. $\endgroup$
    – Sean Eberhard
    Commented Apr 5, 2023 at 17:12

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