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Let $k$ be $\mathbb R$ or $\mathbb C$, and fix some dimension $n$. It is well-known that the orthogonal group $\operatorname O_n(k)$ is a maximal compact subgroup of $\operatorname{GL}_n(k)$, in the sense that any compact subgroup $H\subseteq\operatorname{GL}_n(k)$ is conjugate to some subgroup of $\operatorname O_n(K)$. There is a $p$-adic analogue, for which I refer to Conrad's exposition.

I am interested in an analogue in algebraic geometry, for a general field $k$. More concretely, is the following statement true?

Fix a field $k$. Then there is a maximal proper algebraic subgroup $H$ of the algebraic group $\operatorname{GL}_n$ in the following sense: for any proper algebraic subgroup $H'\subseteq\operatorname{GL}_n$, we can conjugate $H'$ into an algebraic subgroup of $H$.

If this is not true, can one add suitable adjectives (for example, to $k$ or to the subscheme) to make it true?

Derive Foiler
  • 709
  • Up to conjugation it contains $\mathrm{SL}_n$ and because of this one can quickly reduce to the case of $\mathbb{G}_m$. But there the answer is obviously no: the algebraic subgroups are just the $\mu_N$ for $N\geqslant 1$ and these have no largest member. – Alex Youcis Nov 26 '22 at 04:39
  • $O_n(\mathbb C)$ is not compact. – Kenta S Nov 26 '22 at 19:53

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