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Let $G$ be an algebraic group with closed normal subgroup $N$. Suppose that $N$ and $G/N$ are both unipotent. Does it imply that $G$ is also unipotent?

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Yes, see for example Milne's notes AGS, XV, 2.5. –  anon Feb 15 '13 at 0:24
    
@Xingting: It's best to specify whether or not $G$ is connected though it's not an issue in characteristic 0. Anyway, the answer is definitely yes. If you already have the full Borel-Chevalley structure theory of (connected) linear algebraic groups in hand, it's an easy consequence: modulo the unipotent radical of $G$, you get a reductive group (which can't be unipotent if nontrivial). Naturally you might prefer short-cuts. –  Jim Humphreys Feb 15 '13 at 1:06
    
@Jim In order to define the unipotent radical, you need to know enough about unipotent groups to answer the question. And connectedness is not a problem in any characteristic, at least, not if you are talking about algebraic group schemes. –  anon Feb 15 '13 at 1:24
    
What are the implicit smoothness hypotheses on $G$ and $N$? (One doesn't need any such hypotheses for the affirmative answer to the question, but it affects the extent to which one can give an entirely elementary proof or not.) Also, the OP should clarify what they wish to take as the initial definition of "unipotence" (and under what smoothness hypotheses). As many readers here know well, there are multiple different-looking definitions, all ultimately equivalent, but the choice of initial definition affects which basic facts are easy to prove and which require more development. –  user30379 Feb 15 '13 at 2:54
    
Thanks very much for all your answers. –  Xingting Feb 15 '13 at 20:08

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By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, $V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent.

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Your "definition" is not the usual one, since the notion of "unipotent" arises at first in the treatment of Jordan decomposition in a linear algebraic group (over an algebraically closed field of arbitrary characteristic). –  Jim Humphreys Feb 15 '13 at 1:07
    
In Jantzen's RAGS, he asserts that for arbitrary k-group schemes (k a commutative ring), one takes this as the defining characteristic of a unipotent k-group scheme. –  Christopher Drupieski Feb 15 '13 at 1:19
    
There is some variation in the definition in the literature, but the definition I give essentially agrees with that in SGA 3, Demazure Gabriel, Conrad-Gabber-Prasad, and Springer 1998. See XV 4.2 of the notes mentioned in an earlier comment. –  anon Feb 15 '13 at 1:48
    
Dear anon: The definition of unipotence in SGA3 seems to "essentially agree" with yours only in the sense that they're logically equivalent; the proof of the equivalence isn't trivial. To be precise, the definition in C-G-P is taken to be the one in SGA3, which in turn is given in XVII 1.1 (and 1.2): for some (equivalently, any) algebraically closed extension $K$ of $k$, $G_K$ has a composition series whose successive quotients are $K$-subgroups of $\mathbf{G}_{\rm{a}}$ (also see 3.2, 3.5, and 4.1.1(i),(ii) in Expose XVII). –  user30379 Feb 15 '13 at 2:50
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@pranavk In the old days, people had a lot of trouble defining things like "unipotent algebraic group". See Jim's comment, where he requires Borel-Chevalley classification theory to prove that an extension of unipotent algebraic groups is unipotent. It's hard to imagine a more natural definition than that an algebraic group G is unipotent if, for every finite-dimensional representation V, there exists a basis of V such that G acts through $U_n$ (upper triangular matrices with one's on the diagonal). The chapter where unipotent groups are studied, and the defns shown equivalent is 10 pages long –  anon Feb 15 '13 at 3:12

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