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Let $V$ be an $n$-dimensional vector space over a characteristic $0$ field $k$ (or better, let $V=\mathbb{A}^n_k$). I wonder whether the following is true:

Absolutely irreducible subgroups $H$ of $\operatorname{GL}(n)=\operatorname{Aut}(V)$, where the latter two groups are regarded as algebraic groups, are reductive.

By absolute irreducibility I mean the subgroup $H_K$ acts irreducibly on $V_K$ for any field extension $k\to K$, i.e., there is no parabolic subgroup of $\operatorname{GL}(n)$ containing $H$. If it is true, I would like a reference to a proof; if it is not true, I wonder where I can find a counter-example.

My opinion on this is that I need to see whether the set of irreducible groups can inject into the set of reductive groups (or the set of non-reductive ones can inject into the set of non-irreducible ones). On the one hand, there is a correspondence of parabolic subgroups and cocharacters of $\operatorname{GL}(n)$. However, on the other hand, I don't know a classification of reductive subgroups or that of non-reductive ones.

Any help will be appreciated.

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    $\begingroup$ If a connected algebraic group has a faithful absolutely semisimple representation, then it is reductive; see for instance Corollary 22.20 in Milne's notes on algebraic groups. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 30 at 20:51
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    $\begingroup$ @R.vanDobbendeBruyn, isn't that an answer? Anyway, thanks for the neat result, which I didn't know though I thought I knew both Milne's lovely book and complete reducibility fairly well. $\endgroup$
    – LSpice
    Commented Sep 30 at 21:02

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