Automorphism groups of simple groups of Lie type

Question

$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}$In “Automorphisms of finite linear groups”, Steinberg proves that any automorphism of a simple group of Lie type (normal or twisted) is a product of inner-, diagonal-, field-, and graph automorphisms.

Let $G$ be a simple group of Lie type. Denote by $\hat{G}$ the group of automorphisms generated by $G$ and the diagonal automorphisms of $G$, by $\hat{A}$ the group of automorphism generated by $\hat{G}$ and the field automorphisms of $G$, and by $A$ the whole group of automorphisms of $G$.

As Steinberg explains (3.3-3.6 in his paper), we know that the quotient $\hat{G}/G$ is cyclic (there is one exception), the quotient $\hat{A}/\hat{G}$ is cyclic, and the quotient $A/\hat{A}$ is either trivial or has order $2$ or $6$.

All above, I’m fine with. What puzzles me in Steinberg’s work is the order of the quotient $\hat{G}/G$. He says that $\hat{G}/G$ has order $(n+1,q-1)$, $(2,q-1)$, $(2,q-1)$, $(4,q^n -1)$, $(3,q-1)$, $(2,q-1)$, $(n+1,q+1)$, $(4,q^n+1)$ or $(3,q+1)$ for the respective group $A_n$, $B_n$, $C_n$, $D_n$, $E_6$, $E_7$, $A_n^1$, $D_n^1$ or $E_6^1$. I do not understand what he means by this. To my understanding the first number in the brackets is the order of the quotient $\hat{G}/G$ (e.g. in $A_1=\PSL_2(q)$ we have $\hat{G}/G=\PGL_2(F)/{\PSL_2(F)}$ which has order $2=n+1$ when $q$ is odd) but I don’t then understand what the second number in the brackets stands for (e.g. in $A_1=\PSL_2(q)$, what is $q-1$ for? It is the order of the maximal split torus but how is this related to the order of the group $\PGL_2(F)/{\PSL_2(F)}$)?

My second question is about the automorphism groups of Suzuki and Ree groups. They do not appear in Steinberg's work. Is there a similar description for those automorphism groups and where could I find it?

Answer 1

Theorems 30 and 36 in Steinberg's "Lectures on Chevalley Groups," published by the American Mathematical Society, give the automorphism groups of the groups of Lie type over perfect fields. True, the proof of Theorem 36 is only sketched. For the Suzuki and Ree groups, every automorphism is the product of an inner and a field automorphism.

Answer 2

$\DeclareMathOperator\Z{Z}\DeclareMathOperator\C{C}\newcommand\A{\mathsf A}\newcommand\D{\mathsf D}\newcommand\E{\mathsf E}\newcommand\ad{_\text{ad}}\newcommand\sc{_\text{sc}}\newcommand\G{\mathbf G}\newcommand\Gad{\G\ad}\newcommand\Gsc{\G\sc}\newcommand\Fq{\mathbb F_q}\newcommand\Frob{\operatorname{Frob}_q}\newcommand\FFq{\overline{\mathbb F}_q}\newcommand\FFqmult{\smash{\overline{\mathbb F}}_q^\times}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\coker{coker} $I've never quite understood Steinberg's definition of $G$, but I think it may be the image of $\Gsc(\Fq)$ in $\Gad(\Fq)$, with hopefully obvious notation. I'll proceed as if this were so.

I said that $\coker(\Gsc(\Fq) \to \Gad(\Fq))$ was $\Hom(\pi_1(\Gad), \Fq^\times)$ in the split case, but that's not quite true. Instead, I meant to describe the kernel; and in fact, $\ker(\Gsc(\Fq) \to \Gad(\Fq))$ is $\Z(\Gsc)(\Fq) = \Hom_{\mathbb Z}(P/Q, \Fq^\times)$, where $P$ is the integer weight lattice and $Q$ is the integer root lattice, so that $P/Q$ is dual to $\pi_1(\Gad)$. (I thus carelessly double dualised, in some sense.) In general (without assuming splitness), we may say that the kernel is the $\Frob$-fixed point set of $\Hom_{\mathbb Z}(P/Q, \FFqmult)$. In any case, since $\ker(\Gsc(\Fq) \to \Gad(\Fq))$ has the same cardinality as $\coker(\Gsc(\Fq) \to \Gad(\Fq)) = \hat G/G$, it suffices to compute the cardinality of $\Hom_{\mathbb Z}(P/Q, \FFqmult)^{\Frob}$.

As I mentioned in the comments (1 2), if we are in the untwisted case, and either outside of type $\D_n$ or in type $\D_\text{odd}$, then $P/Q$ is cyclic (and carries the trivial action of $\Frob$), so the cardinality of $\Hom_{\mathbb Z}(P/Q, \Fq^\times)$ is $\gcd(\lvert P/Q\rvert, q - 1)$, which equals $\gcd(4, q^n - 1)$ if we are in the $\D_\text{odd}$ case; whereas, if we are in (untwisted) type $\D_\text{even}$, then $P/Q$ is the Klein Vierergruppe (still with the trivial $\Frob$ action), and the cardinality of the hom-set is $\gcd(2, q - 1)^2 = \gcd(4, q^n - 1)$.

In the twisted $\A_n$ and $\E_6$ types, and the twisted $\D_\text{odd}$ type, again $P/Q$ is cyclic, and now $\Frob$ acts on it by inversion. We thus have that $\Hom_{\mathbb Z}(P/Q, \FFqmult)^{\Frob}$ equals $\Hom_{\mathbb Z}(P/Q, \ker \operatorname N_{\mathbb F_{q^2}/\Fq})$, whose cardinality equals $\gcd(\lvert P/Q\rvert, \lvert\ker \operatorname N_{\mathbb F_{q^2}/\Fq}\rvert) = \gcd(\lvert P/Q\rvert, q + 1)$, which again equals $\gcd(4, q^n + 1)$ in type $\D_\text{odd}$, just as before. Finally, in type $\mathsf D_\text{even}^1$, again $P/Q \cong \C_2 \oplus \C_2$, this time with $\Frob$ switching the two summands, so $\Hom_{\mathbb Z}(P/Q, \FFqmult)^{\Frob}$ is isomorphic to $\Hom_{\mathbb Z}(\C_2, \mathbb F_{q^2})$, whose cardinality is $\gcd(2, q^2 - 1) = \gcd(4, q^n + 1)$.