Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$\textbf{Some definitions:}$ Let $G$ be an algebraic group (for me that is the complex points of an affine algebraic group). We say $G$ is reductive if its unipotent radical (maximal connected normal unipotent subgroup) $R_u(G)$ is trivial and we say $G$ is semisimple if its radical (maximal connected normal solvable subgroup) $R(G)$ is trivial. The derived group $\mathcal{D}(G)$ of $G$ is the intersection of the normal subgroups $N$ such that $G/N$ is commutative. If there are algebraic subgroups $H$ and $K$ of $G$ such that the product morphism $H\times K\rightarrow G$ is surjective with finite kernel we say that $G$ is an almost-direct product of $H$ and $K$. (edit: as pointed out in the comments/answers I forgot to specify that the images of $H$ and $K$ commute in $G$.)

$\textbf{A theorem:}$ When $G$ is a connected reductive complex algebraic group, it is the almost-direct product of a central torus $R(G)$ and a semisimple group $\mathcal{D}(G)$. (See page 181 in Borel's book, or page 168 in Humphreys' book "Linear Algebraic Groups".)

$\textbf{My question:}$ I would like to know to which extent this decomposition fails to generalize in the case where $G$ is disconnected. More precisely, I would be interested to know if there are similar decomposition results for a reductive group $G$ (not necessarily connected) that happens to be nilpotent or solvable.

share|improve this question
For "almost-direct product", you mean for $H$ and $K$ to have commuting images in $G$, right? Passing to the quotient by the identity component of $\mathcal{D}(G)$, the problem reduces to an extension $G$ of a finite group $F$ by a torus $T$. The semi-direct product of $\mathbf{Z}/(2)$ against GL$_1$ via inversion (or any finite group against a nontrivial action on a torus) is a counterexample, if I'm not misunderstanding the question. How would you like to rule these out? –  user29283 Apr 25 '13 at 5:47

2 Answers 2

up vote 3 down vote accepted

A solvable reductive group has a very simple structure - the component group is solvable, and the connected component of the identity is an algebraic torus. Unfortunately it is not really possible to decompose it further, because the conjugation action would need to decompose as well, but there are many indecomposable actions of solvable finite groups on tori. These basically boil down to irreducible solvable subgroups of $GL_n(\mathbb Z)$, $n$ the dimension of the torus. There are lots of these.

For nilpotent groups, the situation is a bit better. The conjugation action must be trivial, so you can write the group as the almost direct product of a central torus and a finite group as follows: The torus is the connected component of the identity. The finite group is the kernel of the universal map from the group to a torus.

share|improve this answer
@Will Thank you for the answer, it is along the lines of what I was looking for. I can see why the connected component of the identity in a nilpotent reductive group is a central torus. However, it isn't clear to me what you mean by the "universal map from the group to a torus" which yields the finite group in the almost-direct product decomposition. Could you perhaps elaborate a bit on this? –  Victor Apr 26 '13 at 2:38
Actually my idea doesn't quite make sense, so I'll describe a better way. Consider the map from the group to its abelianization - because the connected component of the identity is clearly central, the kernel of this map is clearly finite. The abelianization is an abelian reductive group, hence it's non-canonically the product of a torus and a finite group. Take the kernel of the composed map from the group to to this torus. That's the finite group you want. –  Will Sawin Apr 26 '13 at 3:11

Maybe it's useful to add a few further clarifications to the original question, which Will has answered for solvable or nilpotent algebraic groups.

1) The basic definitions are the same over any algebraically closed field $K$, though the label "reductive" may be misleading since the group need not act completely reducibly in a finite dimensional representation. But in the three textbooks with the same title (by Borel, Springer, and myself), the theory is developed under the restrictive assumption that $G$ is a connected linear algebraic group. This was the focus of the Chevalley classification and already involves considerable work. When $G$ is allowed to be disconnected (including finite matrix groups), the same basic outline can be followed. But then the extension of the identity component relative to the finite quotient $G/G^\circ$ need not split and is often delicate to sort out.

2) In the definition of almost direct product for an abstract group, Borel (in his preliminaries) specifies a group with two or more normal subgroups, such that the product map is a surjective group homomorphism with finite kernel. In particular, as xuhan comments, the subgroups commute with each other..

3) It's helpful to have in mind some natural examples of disconnected reductive groups. For instance, take $G$ to be connected and semisimple, with a maximal torus $T$ and its normalizer $N$ in $G$. Then $N$ fails to be connected, but has $T$ as identity compoennt and quotient equal to the Weyl group. This example already indicates the tricky nature of disconnected groups, explored by Tits and others: see my references in my earlier answer here. Note that $N$ doesn't need to be solvable or writable as a semidirect product relative to $T$ but does admit an "almost semidirect produuct" decomposition. Another class of examples occurs when $G$ fails to be simply connected; then the centralizer in $G$ of a semisimple element is always reductive but need not be connected. However, the Borel-Tits description of such centralizers shows how to describe the part outside the identity component in terms of a subgroup of the Weyl group.

At any rate, examples of the type just mentioned seldom admit almost direct product decompositions even when solvable.

share|improve this answer
@Jim Thank you for the additional clarifications on the bigger picture. The natural examples you mentioned are indeed quite helpful. –  Victor Apr 27 '13 at 22:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.