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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$Let $K$ be a field of characteristic zero, and $\SL_n$ and $\GL_n$ the special and general linear groups over $K$. Let $\Phi \in \GL_n(K), H \in \SL_n(K)$. Is there always a solution $X \in \SL_n(K)$ to the equation $X\Phi X^{-1}\Phi^{-1} = H$?

If needed, I am happy to assume that $\Phi$ is semisimple and to only ask for a solution in a finite extension of $K$. If you have a reference for general semisimple groups, that would be wonderful.

I believe this goes by the general name of word equations in groups with constants, and I am aware of the theorem that for semisimple $G$ the commutator is the whole of $H$, but I can't seem to find the precise answer to my question in any of the references I've seen.

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    $\begingroup$ $\Phi$ being fixed, the given map factors through a map $\mathrm{GL}_n/C_\Phi\to\mathrm{SL}_n$ and $C_\Phi$ has dimension $\ge n$, so the image has codimension $\ge n-1$. $\endgroup$
    – YCor
    Commented Jan 23, 2023 at 4:01
  • $\begingroup$ I don't understand the downvotes on this question: it's a perfectly legitimate question, and the answer demonstrates that it's not obvious and that a solution was found interesting enough to publish. $\endgroup$
    – Gro-Tsen
    Commented Jan 23, 2023 at 9:35
  • $\begingroup$ @Gro-Tsen this is because the answer is not an answer to the question (which has an immediate negative answer — $\Phi$ being fixed), but to another, related, question. $\endgroup$
    – YCor
    Commented Jan 23, 2023 at 10:31
  • $\begingroup$ (I forgot to write that $C_\Phi$ denotes the centralizer of $\Phi$.) $\endgroup$
    – YCor
    Commented Jan 23, 2023 at 11:15
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    $\begingroup$ $H$ is not a great option for a group element when $H$ is expected to be a subgroup too. $\endgroup$
    – YCor
    Commented Jan 23, 2023 at 14:15

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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$EDIT: The below answers a slightly different question of whether a matrix in $\SL(n,K)$ is a commutator. The question by OP asks whether for fixed $\Phi$, every matrix in $\SL(n,K)$ is a commutator $\Phi A \Phi^{-1} A^{-1}$. The comment by YCor shows that the answer is negative.

Let $K$ be a field.

The paper "Thompson, R. C., Commutators in the special and general linear groups, Trans. Amer. Math. Soc. 101 (1961), 16–33." (link) claims the following theorems, quote:

Theorem 1. Let $\rho I_n \in \SL(n, K)$. Then $\rho I_n$ is always a commutator of $\GL(n, K)$. Moreover, $\rho I_n$ is a commutator of $\SL(n, K)$ unless $\rho$ is a primitive $n$th root of unity in $K$ and $n \equiv 2 \mod{4}$.

In this exceptional case $\rho I_n$ can always be expressed as a product of two commutators of $\SL(n, K)$ and can be expressed as a single commutator of $\SL(n, K)$ when, and only when, the equation $-1 =x^2+y^2$ possesses a solution $x,y \in K$. This condition is always satisfied when $K$ has characteristic different from zero.

Theorem 2. Let $A \in \SL(n, K)$. If $A$ is not scalar and if $K$ has at least four elements, then $A$ is a commutator of $\SL(n, K)$.

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