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Suppose $G$ is a connected Lie group whose radical is $R$. It is known that the solvable group $R$ can always be decomposed as $R=UT$ where $U$ is a simply-connected normal subgroup of $R$ and $T$ is a compact abelian subgroup of $R$ with $U\cap T = 1_G$. We know that $R$ is a normal subgroup of $G$.

Question: Is $U$ necessarily a normal subgroup of $G$?

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    $\begingroup$ Your statement "it is known..." is not true. Example: $G=R$ the quotient of the 3-dimensional Heisenberg group by an infinite discrete central subgroup. $\endgroup$
    – YCor
    Commented Mar 25, 2013 at 17:40
  • $\begingroup$ Even assuming that it holds (a sufficient condition being that $G$ is linear; a weaker one being that $R$ is linear), the question is unclear. Indeed, there exists a decomposition $UT$... is $U$ normal in $G$? normal for every such decomposition? for at least one decomposition? $U$ is not unique in general. $\endgroup$
    – YCor
    Commented Apr 1, 2019 at 7:36

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In the case of connected linear algebraic groups it is true: Any inner automorphism of $G$ is an algebraic group automorphism of $R$. And so it carries all the unipotent elements of $R$ to unipotent elements, (See Section 19. Connected Solvable Groups in J.E. Humphreys textbook "Linear Algebraic Groups")

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  • $\begingroup$ In general, $U$ is not the unipotent radical. For instance, if $G$ is the affine group $\mathbf{R}\ltimes\mathbf{R}$, then $U=G$. $\endgroup$
    – YCor
    Commented Apr 1, 2019 at 7:38
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As Yves Cornulier already said: Your presumed statement is wrong.

Any connected, linear, solvable Lie group over the reals is the semi-direct product of a compact abelian subgroup and a simply connected normal subgroup. (This holds more general for algebraic, connected, solvable lie groups ober a field of characteristic 0, what can be found in Chevalley's "Théorie des groupes de Lie")

This is in general false for non-linear Lie groups, what explains Yves Cornulier's counter example as the quotient of the heisenberg group with it's central discrete cyclic subgroup is not linear.

(The non-linearity of this group is e.g. proved in "The Structure of Compact Groups: A Primer for Students, a Handbook for the Expert" of Hofmann and Morris on page 169.)

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  • $\begingroup$ Thank you. But if G is linear, the answer to the question is Yes? $\endgroup$
    – Li Yu
    Commented Mar 26, 2013 at 2:45
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    $\begingroup$ @LiYu: you should edit your question accordingly. $\endgroup$
    – YCor
    Commented Apr 30, 2014 at 10:27

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