# Are all connected solvable affine algebraic groups supersolvable?

The basic question is whether there is a notion of chief factor of a connected solvable algebraic group that matches my intuition. A few smaller assertions are sprinkled through the explanation, and the implicit question is if these are correct (are unipotent groups nilpotent, are their chief factors all isomorphic to subgroups of Ga, etc.).

In J.S. Milne's course notes on the basic theory of algebraic groups, theorem 14.30:

A connected solvable smooth group over a perfect field has a connected unipotent normal subgroup whose quotient is of multiplicative type.

In other words, the derived subgroup is contained in the unipotent radical.

Now, a connected group acts on a group of multiplicative type trivially, by 13.21.

I did not see it mentioned, but I believe unipotent groups are nilpotent in both the group theoretic sense and whatever fancy definition one might cook up for these functors. I think that the lower central factors should be direct products of subgroups of the additive group Ga.

By the Jordan decomposition or 13.13 or 13.15, I think any action of Gm on (Ga)^n is diagonal.

It looks like a connected solvable affine algebraic group over an algebraically closed field has a chief series consisting of subgroups of Ga and Gm, all of which I would describe as being one-dimensional.

The analogy with finite groups takes the unipotent radical to be O_p(G), the p-core, and so it appears that a finite group of solvable algebraic type always has [G,G] <= O_p(G), so that not only is G supersolvable nilpotent-by-abelian, it is p-closed and its eccentric chief factors are all for the same prime p. In the algebraic case, the central chief factors would be the subgroups of Gm, and the eccentric chief factors would be the subgroups of Ga.

In other words, connected solvable affine algebraic groups over algebraically closed fields are very dissimilar from finite solvable groups in that the representations they define on their own chief factors are all one-dimensional. More briefly, connected solvable affine algebraic groups over algebraically closed fields are supersolvable.

Edit: I think the answer to my question is relatively simple: "supersolvable" is a little tricky to directly generalize, but "nilpotent-by-abelian" is quite easy and true, and still implies that any chief factors will be one-dimensional. In Jim's answer, it appears Borel-Serre-Mostow (at least by the time they are translated into Russian) also considered these groups to be "supersolvable", so the name is reasonable, even if the correct definition is just "nilpotent-by-abelian".

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What's your base field? The answer to "are the chief factors [of a unipotent group] all isomorphic to Ga" is "no" if you'll allow me a wacky base field (characteristic p with inseparable extensions). – Kevin Buzzard Mar 31 '10 at 21:51
I'm happy with whatever nice field assumptions one wants as long as it allows thinking about finite groups. I assume something like "let the field be the algebraic closure of its prime field" is general enough for me. The only examples I've heard of where the chief factors of a unipotent group are not exactly Ga have the property that they are subgroups of Ga, and so, I think, are still one-dimensional. It seems similar to how groups of multiplicative type don't need to be Gm, but are subgroups of Gm. – Jack Schmidt Mar 31 '10 at 22:49
Edited to reflect more of the intent (added "over an alebraically closed field" and "subgroups of"). Basically there is an intuition that the action of a solvable group in this fancy sense on its abelian normal subquotients is fairly trivial compared the action of a solvable group in the group theoretic sense. For instance, there is nothing like S4 = AGL(2,2) where the unipotent radical, K4, does not contain the derived subgroup A4. Since it is a natural matrix group, this is surprising to me. However, it exists precisely because it violates Lie–Kolchin's theorem. – Jack Schmidt Mar 31 '10 at 23:06

Edit: According to Wikipedia, $A_4$ is not supersolvable, but $A_4 \cong (\mathbb{G}_a \rtimes \mathbb{G}_m)(\mathbb{F}_4)$. I think this answers your question in the negative.