# Automorphism groups of simple groups of Lie type

$$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}$$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?

• Note that the order of $\operatorname{PGL}_2(F)/{\operatorname{PSL}_2(F)}$ is $2$ when $q$ is odd and $1$ when $q$ is even: exactly the greatest common divisor $(2, q - 1)$, as you say. For the un-twisted non-$\mathsf D_n$ types, you are measuring the homomorphisms from the fundamental group, which is cyclic of order (let's say) $f$, to $\mathbb F_q^\times$, and there are $(f, q - 1)$ such homomorphisms. Sep 22, 2021 at 7:14
• The formula for $\mathsf D_n$ seems to me to be just a coincidentally pleasantly concise way of avoiding an explicit parity distinction. If $n$ is odd, then the same works for $\mathsf D_n$, where $(f, q - 1) = (4, q - 1) = (4, q^n - 1)$. If $n$ is even, then there are $(2, q - 1)^2 = (4, q^n - 1)$ such homomorphisms. I suspect the same kind of explanation works in the twisted case, but I'm not very familiar with it. Sep 22, 2021 at 7:21
• @LSpice, thank you! So you are saying that the order or $\hat{G}/G$ is the greatest common divisor of the pair in the brackets? This makes sense (e.g. in $PGL_2(F)/PSL_2(F)$ as you explained). Somehow it was not explained in Steinberg paper that he talks about the greatest common divisor - or perhaps I missed it. Do you happen to know a reference for automorphisms groups of simple groups of Lie type other that Steinbergs paper? Sep 22, 2021 at 7:36
• The notations $(a, b)$ for $\operatorname{gcd}(a, b)$ and $[a, b]$ for $\operatorname{lcm}(a, b)$ are common in some contexts, so Steinberg may have assumed that it would be clear to the reader. I'm afraid I don't know any other references. Sep 22, 2021 at 13:35

$$\DeclareMathOperator\Z{Z}\DeclareMathOperator\C{C}\newcommand\A{\mathsf A}\newcommand\D{\mathsf D}\newcommand\E{\mathsf E}\newcommand\ad{_\text{ad}}\newcommand\sc{_\text{sc}}\newcommand\G{\mathbf G}\newcommand\Gad{\G\ad}\newcommand\Gsc{\G\sc}\newcommand\Fq{\mathbb F_q}\newcommand\Frob{\operatorname{Frob}_q}\newcommand\FFq{\overline{\mathbb F}_q}\newcommand\FFqmult{\smash{\overline{\mathbb F}}_q^\times}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\coker{coker}$$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 $$\gcd(2, q^2 - 1) = \gcd(4, q^n + 1)$$.
• The explanation is so pleasant away from type $\mathsf D$ that I'm tempted to think it's the right one, but so unpleasant in type $\mathsf D$ that I'm tempted to think it's the wrong one. There are people here with much more expertise than I about finite groups of Lie type, and hopefully one of them will come along to explain a better perspective. Sep 22, 2021 at 23:19