Automorphism group of the affine groups AGL(n,q), ASL(n,q)

Question

I have a question. The automorphism group of the linear groups $GL(n,q)$, the group of linear transformations of $V = \mathbb{F}_q^n$, and $SL(n,q)$, the subgroup of $GL(n,q)$ consisting of elements of determinant $1$, are well understood. Here $q$ is a prime power $p^r$ and $n$ is any positive integer.

My question is: what is known about the automorphism group of $AGL(n,q)$ and $ASL(n,q)$? Here $AGL(n,q)$ is the affine linear group $V \rtimes GL(n,q)$, semidirect product of $V = \mathbb{F}_q^n$ and $GL(n,q)$ with the natural action of $GL(n,q)$ on $V$. These groups have matrix representations, as you can see here https://en.wikipedia.org/wiki/Affine_group

What I know for sure is that there are the inner automorphisms, and the Frobenius automorphism (raise the coefficients to the power $p$) that comes from the matrix representation described in the link. But is there something else?

Is there a nice complete description of the automorphism groups of $AGL(n,q)$ and $ASL(n,q)$?

Thanks

Answer 1

This is just a brief answer. The only autmorphisms other than the ones that you know about already arise from elements of $H^1(H,V)$ (with $H = {\rm SL}(n,q)$ or ${\rm GL}(n,q)$), where the corresponding automorphisms induce the identity on $V$ and on $G/V \cong H$, but map a complement $H$ of $V$ to a complement that is not conjugate to $H$ in $G$.

These cohomology groups are all known. The only cases in which they are non-zero are when $H = {\rm SL}(2,2^k)$ with $k \ge 2$, and $H = {\rm SL}(3,2) = {\rm GL}(3,2)$, and in each of these cases, the dimension of $H^1(H,V)$ is $1$-dimensional over ${\mathbb F}_q$.

So, for example, for $G = {\rm ASL}(2,8)$, the order of the automorphism group is $24|{\rm AGL}(2,8)| = 168|G|$, with a factor $3$ coming from the field automorphisms of ${\mathbb F}_8$, and the factor $8$ coming from $H^1(H,V)$.

As a quick justification, note that $V$ is clearly characteristic in $G$, so ${\rm Aut}(G)$ induces a subgroup of the automorphism group of $G/V$. The full automorphism group of ${\rm SL}(n,q)$ is $\Gamma L(n,q)(.2)$, where the $.2$ is the duality automorphism when $n \ge 3$. The duality automorphism of $G/V$ does not lift to an automorphism of $G$, but we do have $A \Gamma L(n,q) \le {\rm Aut}(G)$.

Since $G$ acts absolutely irreducibly on $V$, automorphisms of $G$ inducing the idenity on $G/V$ must act as scalars on $V$, and we already have the full group of scalars present in $A \Gamma L(n,q)$.

So it remains to consider automorphisms of $G$ that induce the identity on both $G/V$ and on $V$ and, modulo inner automorphisms induced by conjugation by elements of $V$, these correspond exactly to $H^1(H,V)$.

A reference for the results on $H^1(H,V)$ is: G.W. Bell, On the Cohomology of the Finite Special Linear Groups, I, Journal of Algebra 54, 216-238 (1978), but I am not sure whether that was the first proof.

Answer 2

Note that $V = F^{\ast}({\rm AGL}(n,q))$, where $F^{\ast}$ denotes the generalized Fitting subgroup.

For any finite group $G$, restriction gives a homomorphism from ${\rm Aut}(G)$ into ${\rm Aut}(F^{\ast}(G))$ whose kernel $A$ is an Abelian normal subgroup of ${\rm Aut}(G)$ and all prime divisors of $|A|$ are divisors of $|G|$. This is a standard application of the three subgroups lemma, and properties of coprime automorphisms, but I outline the argument. Let $A$ denote the kernel.

Now we have $[G,F^{\ast}(G),A] = [F^{\ast}(G),A,G] = 1$, so that $[A,G,F^{\ast}(G)] =1$. Hence $[G,A] \leq Z(F(G))$, so that $[A,G,A] = [G,A,A] = 1$ Hence $[A,A,G] = 1$, so that $[A,A]$ is trivial. If $|A|$ has a prime divisor $p$ which does not divide $|G|$, and $P$ is a Sylow $p$-subgroup of $A$ then $G = [G,P]C_{G}(P)$ and $[G,P,P] = [G,P]$ while we saw above that $[G,P,P] = 1$. Hence $P$ is trivial.