# The normalizer of a reductive subgroup

Let $k$ be a field and $G$ a linear algebraic group over $k$. Let $H$ be a diagonalizable subgroup of $G$. Then it is a classical fact that the centralizer $C_G(H)$ of $H$ is of finite index in the normalizer $N_G(H)$ of $H$.

Now let $H$ be an arbitrary reductive subgroup of $G$. Is it true that $Z_G(H)\cdot H$ is of finite index in $N_G(H)$?

In the case of certain Lie groups, the answer is yes, see, e.g., D. Poguntke, Normalizers and centralizers of reductive subgroups of almost connected Lie groups (MR1650341), which motivates my question.

-
If you do not require $H$ to be connected then it becomes tricky. Automorphisms of non-connected reductive groups may be worth a separate question. – Wilberd van der Kallen Jun 7 '10 at 15:47

## 2 Answers

While the comments by Angelo and Wilberd go a long way toward an answer, I'd prefer to start with a more precise formulation of the question. The group $G$ and its (closed!) subgroup $H$ may be assumed to be connected with $H$ moreover reductive. The structure theory involved here is essentially independent of the characteristic of the ground field, but on the other hand a field of definition apparently plays no role and could be assumed to be algebraically closed.

Up to isogeny, a connected reductive group is a direct product of closed connected subgroups which can be tori or products of (quasi-)simple groups of a fixed type. Due to the standard theorem on "rigidity" of tori, the centralizer of a torus in $G$ has finite index in the normalizer. On the other hand, the outer automorphism group of a simple group is finite (coming from the automorphisms of the Dynkin diagram) while other outer automorphisms of a direct product of groups of the same type come from just finitely many permutations of isomorphic simple factors.

Since an automorphism of $H$ preserves the center and the various products of simple factors of the same type (whose overall product is the derived group of $H$), these pieces combine to show that $N_G(H)/H \, Z_G(H)$ is finite. (Though I haven't written it all down myself.) I don't recall a specific reference for such a general result in the literature; it may only come up in special situations.

-
You want outer automorphisms, so the $Z_G(H)\cdot H$ in the question is right. – Wilberd van der Kallen Jun 7 '10 at 14:54
Yes, I forgot to put in the factor $H$. Thanks. – Jim Humphreys Jun 7 '10 at 15:55
"Up to finite index"—certainly the identity component of $H$ is of finite index in $H$, and similarly in $G$, but is there any reason for $\mathrm N_G(H)$ to have finite index in $\mathrm N_G(H^\circ)$, or for $\mathrm C_G(H)$ to have finite index in $\mathrm C_G(H^\circ)$? (In fact, this latter can certainly fail.) Maybe I misunderstood the reduction, though. – L Spice Jun 9 at 23:05
@L Spice: Yes, this was written down without editing, though at the time I probably intended to go back and finish it up. I've clarified a bit, but the question is probably dead by now. – Jim Humphreys Jun 11 at 15:28

I think so. This is implied by the fact that the group of outer automorphisms of $H$ is finite. When $H$ is semisimple, then the group of outer automorphisms is contained in the group of automorphisms of the Dynkin diagram, so it is finite. The general case should reduce to this by considering the commutator subgroup of $H$, which is semisimple.

-
As the center of $H$ is also preserved by automorphisms of $H$, one may as well consider the product of this center with the commutator subgroup of $H$. Now the image of $N_G(H)$ in the outer automorphisms of the product is still finite. And the product has finite index in $H$. – Wilberd van der Kallen Jun 7 '10 at 10:23
What I should have said is that the direct product maps onto $H$. The finite index in $H$ is thus actually 1. Thus while the group of outer automorphisms of $H$ need not be finite, the image of $N_G(H)$ in this group of outer automorphisms is finite. – Wilberd van der Kallen Jun 7 '10 at 12:04