Automorphisms of the affine semilinear group $A\Gamma L(1,2^{n})$

Question

In this question, it is mentionned that the group of automorphisms of the semilinear group $A\Gamma L(1,2^{n})$ is the group itself.

Do you have a short proof of this fact?

Answer 1

$G={\rm AGL}(1,q)$ is the group of transformations of $K={\mathbb F}_q$ of the form $x \to ax+b$ with $a \in K$, $0 \ne b \in K$. It is generated by the translations $\tau_a:x \to x+a$ and the multiplications $\mu_b:x \to bx$. Note that $\mu_b\tau_a\mu_b^{-1}=\tau_{ab}$.

Let $\phi$ be an automorphism of $G$. The translation group is characteristic (it is a normal Sylow subgroup of $G$) and hence is fixed by $\phi$. So we can write $\phi(\tau_a) = \tau_{\psi(a)}$ where $\psi$ as is a permutation of $K$ induced by $\phi$. Since $\phi$ fixes the identity of $G$, we have $\psi(0)=0$. Also, by composing $\phi$ with the inner automorphism induced by $\mu_b$ for some $b$, we can assume that $\psi(1)=1$.

We claim that $\psi$ is an automorphism of $K$, which will prove that ${\rm Aut}(G)={\rm A \Gamma L}(1,q)$.

$\psi(a_1+a_2)=\psi(a_1) + \psi(a_2)$ follows from $\phi(\tau_{a_1}\tau_{a_2}) = \phi(\tau_{a_1})\phi(\tau_{a_2})$.

Now, since $\psi(1)=1$, $\phi$ must fix the stabilizer in $G$ of $1 \in K$, which the multiplication group of elements $\mu_b \in G$. Then $$\phi(\mu_b)\tau_1\phi(\mu_b^{-1}) = \phi(\mu_b)\phi(\tau_1)\phi(\mu_b^{-1}) = \phi(\mu_b\tau_1\mu_b^{-1}) = \phi(\tau_{b}) = \tau_{\psi(b)}=\mu_{\psi(b)}\tau_1\mu_{\psi(b)}^{-1}$$ implies that $\phi(\mu_b) = \mu_{\psi(b)}$, and then $\psi(b_1b_1)=\psi(b_1)\psi(b_2)$ follows from $\phi(\mu_{b_1}\mu_{b_2}) = \phi(\mu_{b_1})\phi(\mu_{b_2})$.