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Let $G<\mathrm{SL}_n$ be a linear algebraic group defined over a field. Is there a bound on the number of connected components of $G$ in terms of $n$ alone?

(The bound will evidently not be any smaller than $n!$. It is also clear why we are requiring $G<\mathrm{SL}_n$ and not $G<\mathrm{GL}_n$: the variety $\det(g)^k-1$ has $k$ connected components with each connected component being of the form $\det(g) = \omega$, $\omega$ a $k$th degree of unity. Clearly we cannot bound $k$ in terms of $n$.)

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    $\begingroup$ Maybe I don't understand the definition of 'linear algebraic group'. It seems to me that the subgroup $G_k\subset\mathrm{SL}(2,\mathbb{C})$ defined by the equations $g\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}g$ and $g^k=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has at least $k$ components. Is this $G_k$ not 'linear algebraic'? $\endgroup$
    – Robert Bryant
    Commented Aug 15, 2023 at 12:29
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    $\begingroup$ Maybe you want to work modulo abelian normal subgroups, as in Jordan's Theorem and its variants. $\endgroup$
    – Dave Benson
    Commented Aug 15, 2023 at 13:04
  • $\begingroup$ @RobertBryant Right, that's the general sort of thing I wanted to avoid (roots of unity) and your example makes it clear that requiring $G<\mathrm{SL}_n$ is not enough. Thanks. $\endgroup$
    – H A Helfgott
    Commented Aug 15, 2023 at 13:37
  • $\begingroup$ @DaveBenson Well, Jordan's theorem is not true in characteristic $\ne 0$, at least not without additional conditions. Larsen-Pink amounts to a version that works for arbitrary characteristic. I suppose one cannot really do better than that? $\endgroup$
    – H A Helfgott
    Commented Aug 15, 2023 at 13:44
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    $\begingroup$ It might be worth noting that $GL_n$ embeds into $SL_{n+1}$ via $g \to \begin{pmatrix}g&0\\0&det(g)^{-1}\end{pmatrix}$, so your $GL$ example is also an $SL$ example. $\endgroup$
    – Nate
    Commented Aug 15, 2023 at 18:38

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