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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\PSL{PSL}$Consider the general linear group $\GL(n,q)$ over a finite field with $q$ elements and the corresponding special linear group $\SL(n,q).$

Consider the group $\PGL(n,q)=\GL(n,q)/Z_{\GL(n,q)}$ and its subgroup $\PSL(n,q)=\SL(n,q)/{Z_{\SL(n,q)}}\cong (\SL(n,q)\cdot Z_{\GL(n,q)})/{Z_{\GL(n,q)}},$ where $Z_G$ denotes the center of the group $G$.

The following relations for the centralizer seems to be well-known $$C_{\GL(n,q)}(\SL(n,q))=Z_{\GL(n,q)}$$ and $$C_{\PGL(n,q)}(\PSL(n,q))=\{I\}$$ (they can be proved easily once one knows that $\SL$ is generated by transvections, actually I think they are true over more general rings) but I am unable to find a source for these facts.

Can you provide a book/paper where I can find them?

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    $\begingroup$ Would it suffice to reduce to the observation that $Z_{\operatorname{GL}(n, Q)} \cap \operatorname{GL}(n, q) = Z_{\operatorname{GL}(n, q)}$, and that the centre of $\operatorname{PGL}(n, Q)$ is trivial, for every power $q$ of $Q$ (and give a citation for that) by observing that $\operatorname{GL}(n, q)$ is contained in $\operatorname{SL}(n, q)\cdot Z_{\operatorname{GL}(n, Q)}$ for some high power $Q$ of $q$, and analogously for $\operatorname{PGL}(n, q)$? $\endgroup$
    – LSpice
    Commented Oct 23 at 2:04
  • $\begingroup$ Could you elaborate on that filling the details? $\endgroup$
    – Nick Belane
    Commented Oct 23 at 15:47
  • $\begingroup$ Re, ah, I'm sorry, I am so used to thinking of algebraic groups that I didn't notice that I was base-changing twice, not just once; that is, $\operatorname{GL}(n, q)$ is contained in $\operatorname{SL}(n, Q)\cdot Z_{\operatorname{GL}(n, Q)}$, not necessarily in $\operatorname{SL}(n, q)\cdot Z_{\operatorname{GL}(n, Q)}$. $\endgroup$
    – LSpice
    Commented Oct 23 at 19:00

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