# Connected extensions of finite by connected algebraic groups

Let $1\to H\to E\to G\to 1$ be a short exact sequence of algebraic groups defined over an algebraically closed field $k$ of characteristic $p$. Suppose $H$ is a finite group, and $G$ and $E$ are connected. Does it follow that $G\cong E$?

[Edit: of course not in general, since, as Max points out, E=SL_n, H=Z(E) and G=PSL_n yield counterexamples]

More specifically for my purposes, if $G$ is a vector group, (i.e. $G$ is isomorphic to the direct product of $n$ copies of the additive group $G_a(k)$ of the field $k$) must $E$ be one as well? (Let $E$ also be unipotent, if it helps.)

I note that this follows when $G=G_a(k)$, the additive group of the field, since there are only two connected 1-dimensional groups and $G_m(k)$ does not surject onto $G_a(k)$.

Differentiating, one sees that $L(E)\cong L(G)$, for what it's worth.

In Groupes algébriques et corps de classes Serre classifies the $2$-dimensional commutative unipotent connected algebraic groups $G$ (VII:11). With the exception of the product of the additive group with itself they are all isogenous to the Witt vector group $W_2$ so that there is an exact sequence as per above with $E=W_2$. For some of them Serre notes that the isogeny can be chosen to be separable so that $H$ is a finite group (i.e., étale group scheme) yet $G$ is not necessarily isomorphic to $W_2$.
Addendum: An algebraic group is a vector group precisely when it is commutative, connected and killed by $p$ (again in Serre's book). Clearly $G$ is killed by $p$ if $E$ is. Conversely, if $G$ is killed by $p$ then multiplication by $p$ induces a map $G\to H/pH$ which is constant as $G$ is connected and $H$ is finite. Thus $G$ is a vector group precisely when $E$ is. Also vector groups are determined by their dimension and $E$ and $G$ have the same dimension.
• I suppose it is still possible that if $G$ is a vector group, then $E$ will be also. Aug 26, 2011 at 13:37
• A vector group is an algebraic group isomorphic to a direct product of n copies of $G_a(k)$. I suppose it's distinct from a vector space in the sense that there's no assumed action of $k^*$ on it. Aug 26, 2011 at 13:54