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I am trying to read the article "Three-dimensional affine crystallographic groups" of Fried–Goldman (Adv. Math., 1983). At some place, it states that if $G$ is a connected solvable closed subgroup of $\operatorname{GL}(\mathbb R^n)$, then its normalizer in $\operatorname{GL}(\mathbb R^n)$ is an algebraic group. I couldn't see how this works. I would like to know if this is a classical fact, and if there is any reference on that ?

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    $\begingroup$ The normalizer of every connected closed subgroup is the stabilizer of its Lie algebra, so is Zariski-closed. $\endgroup$
    – YCor
    Dec 1, 2023 at 23:03
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    $\begingroup$ This seems a reasonable question for someone who is coming to this paper from the point of view of analysis / diff geom but who is not so familiar with Lie theory $\endgroup$
    – Yemon Choi
    Dec 2, 2023 at 12:42
  • $\begingroup$ @YCor Yes I see now, thank you ! $\endgroup$
    – LeeM
    Dec 2, 2023 at 12:51
  • $\begingroup$ Your title mentions that $G$ is linear, but your body does not. Is that intentional? (The normaliser of any (Zariski-closed) linear subgroup of a linear group is linear.) $\endgroup$
    – LSpice
    Dec 2, 2023 at 14:33
  • $\begingroup$ @LSpice the body says that $G$ is in $\mathrm{GL}_n$ and the normalizer is the normalizer in $\mathrm{GL}_n$. $\endgroup$
    – YCor
    Dec 2, 2023 at 14:39

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The normalizer of every connected closed subgroup of $\mathrm{GL}_n(\mathbf{R})$ is the stabilizer of its Lie algebra, so is Zariski-closed.

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