Let G be a simple algebraic group. Let H be a reductive subgroup of G which contains a regular unipotent element of G. Such subgroups were classified by [Saxl and Seitz](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/S0024610797004808) in all good characteristics. I'm actually interested in the characteristic zero version of this result, which apparently goes back to Dynkin. Saxl–Seitz and Dynkin are difficult to read.


I'm wondering if there exists a modern reference for this classification over complex numbers. Or better, could some one please provide the sketch of an argument?


For quick reference, the classification is stated just after Proposition 8 in this [paper](https://arxiv.org/pdf/0901.2163.pdf).



Subgroups of algebraic groups containing regular unipotent elements
I do not know a reference, but I have thought about the same question. Here is a sketch using arguments that are in the literature. For some basics about regular unipotent elements, see for example Chapter 4 of [1]. First a reminder:



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> $(\*)$ Let $\Phi^+$ be a system of positive roots on $\Phi$. A unipotent element $\prod\_{\alpha \in \Phi^+} x\_{\alpha}(c\_{\alpha})$ is regular if and only if $c\_{\alpha} \neq 0$ for all simple roots $\alpha$.
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Let $G$ be simple algebraic group over $\mathbb{C}$ and let $H < G$ be a reductive subgroup containing a regular unipotent element $u \in G$. Denote the root system of $G$ by $\Phi$, with respect to some maximal torus $T$.


Any power of $u$ is also a regular unipotent element and $H/H^\circ$ is finite, so we might as well assume that $H$ is connected. Also $H$ must be semisimple, since $C\_G(u)^\circ$ is unipotent and so $u$ is not centralized by any non-trivial torus.


So $H = H\_1 \cdots H\_t$ is a central product of some simple algebraic groups $H\_i$. Write $u = u\_1 \cdots u\_t$, for some unipotent $u\_i \in H\_i$. The subgroup generated by $u\_i$ lies in the unipotent radical of $H$, so we can assume that the $u\_i$ are contained in $U = \prod\_{\alpha \in \Phi^+} U\_{\alpha}$ with respect to some system of positive roots $\Phi^+$.


Suppose that $t \geq 2$. Then none of the $u\_i$ can be regular in $G$, since they are centralized by $H\_j$ for $j \neq i$. Since $u$ is regular and commutes with $u\_i$, it follows from the lemma below that each $u\_i$ is contained in $\prod\_{\alpha \in \Phi^+ \setminus \Delta} U\_{\alpha}$, where $\Delta$ is the set of simple roots. But then $u \in \prod\_{\alpha \in \Phi^+ \setminus \Delta} U\_{\alpha}$, which contradicts $(\*)$.



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> **Lemma:** Let $u, u' \in U$. Suppose that $u$ is regular and $uu' = u'u$. Then either $u'$ is regular or $u' \in \prod\_{\alpha \in \Phi^+ \setminus \Delta} U\_{\alpha}$.
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> **Proof:** Write $u = \prod\_{\alpha \in \Phi^+} x\_{\alpha}(c\_{\alpha})$ and $u' = \prod\_{\alpha \in \Phi^+} x\_{\alpha}(c\_{\alpha}')$. Then by the Chevalley commutator formula $[u,u'] = \prod\_{\alpha \in \Phi^+ \setminus \Delta} x\_{\alpha}(t\_{\alpha})$ for some $t\_{\alpha} \in \mathbb{C}$.
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> Suppose that $[u,u'] = 1$.
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> We show that if $c\_{\alpha}' \neq 0$ for some $\alpha \in \Delta$, then $c\_{\beta}' \neq 0$ for any $\beta \in \Delta$ adjacent to $\alpha$ in the Dynkin diagram. To this end, from the Chevalley commutator formula we see that $$t\_{\alpha+\beta} = \pm (c\_{\alpha}c\_{\beta}' + c\_{\alpha}'c\_{\beta}).$$ So the claim follows since $c\_{\alpha},c\_{\beta} \neq 0$ by $(\*)$.
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> The Dynkin diagram of $G$ is connected, so the conclusion from this is that either $c\_{\alpha}' = 0$ for all $\alpha \in \Delta$, or $c\_{\alpha}' \neq 0$ for all $\alpha \in \Delta$. By $(\*)$ the latter is same as being regular, so the lemma follows.
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The above lemma is Lemma 2.4 in [2].


In any case, $H$ must be simple. Furthermore, we can show that $u$ must be regular in $H$.



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> **Lemma:** $u$ is regular in $H$.
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> **Proof:** If $u$ is not regular in $H$, then by $(\*)$ there exists a non-Borel parabolic subgroup $P\_H < H$ such that $u \in R\_u(P\_H)$. By the Borel-Tits theorem, there exists a parabolic subgroup $P\_G$ of $G$ such that $P\_H < P\_G$ and $R\_u(P\_H) < R\_u(P\_G)$. But then $P\_G$ is a non-Borel parabolic of $G$ such that $u \in R\_u(P\_G)$, which is a contradiction by $(\*)$.
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By results of Jacobson-Morozov and Kostant, in $G$ there is always a simple subgroup $H$ of type $A\_1$ which contains a regular unipotent element of $G$. Such a $H$ is unique up to conjugacy in $G$, I will call it a ``regular $A\_1$-subgroup''.


Then we still need to consider the case where $H$ is simple of rank $\geq 2$. I won't go through all the details, but at this point we can use some representation theory. Let $X < H$ be regular $A\_1$-subgroup of $H$ (hence of $G$). The idea is that we look at the action of $X$ on a small $G$-module $V$, and this will place a heavy restriction on what $H$ and the $H$-module $V \downarrow H$ can be.


For $G$ of classical type you can take $V$ to be the natural module. In this case $V \downarrow X$ is irreducible if $G$ is not of type $D\_n$, and $V \downarrow X = V\_X(0) \oplus V\_X(2n-2)$ in type $D\_n$.


For types $G\_2$, $F\_4$, $E\_6$, $E\_7$, $E\_8$ take $V$ to be an irreducible $G$-module of dimension $7$, $26$, $27$, $56$, $248$ respectively. You can compute the composition factors of $V \downarrow X$ or look at tables in the literature.


Anyway, in all cases $V \downarrow X$ is multiplicity-free with not too many composition factors. The irreducible $H$-modules $W$ such that $W \downarrow X$ is multiplicity-free have been classified in [3]. You don't need this full result since our situation is even more specific, but the basic technique in the beginning of the paper might be helpful for you.


For example, for $c$ the highest weight of $V \downarrow X$, there is no composition factor of highest weight $c-2$ or $c-4$ (so weights $c$, $c-2$, $c-4$ occur with multiplicity one). From this you can already see that in the restriction $V \downarrow H$, the only possible composition factors are $V\_H(\varpi\_{\alpha})$ (fundamental highest weight corresponding to $\alpha \in \Delta$), where $\alpha \in \Delta$ is an end node of the Dynkin diagram.


Then with similar arguments you can rule out all configurations except those that actually occur (for example $H = F\_4$, $G = E\_6$, with $V \downarrow H = V\_H(\varpi\_4) \oplus V\_H(0)$).




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[1] Humphreys, James E.: Conjugacy classes in semisimple algebraic groups. *Mathematical Surveys and Monographs*, **43**. American Mathematical Society, Providence, RI, 1995.


[2] Testerman, Donna; Zalesski, Alexandre: Irreducibility in algebraic groups and regular unipotent elements. *Proc. Amer. Math. Soc.* **141** (2013), no. 1, 13–28.


[3] Liebeck, Martin W.; Seitz, Gary M.; Testerman, Donna M.: Distinguished unipotent elements and multiplicity-free subgroups of simple algebraic groups. *Pacific J. Math*. **279** (2015), no. 1-2, 357–382.


