Affine Group

Definition

Let $V={\mathbb{F}}_q^n$ be a vector space. It can be viewed as an elementary abelian group $(V,+)$. Then the action of $\mathrm{GL}(V)$ on $V$ induces a group extension:

$${\mathrm{AGL}}_n(q):={\mathbb{Z}}_p^{fn}:{\mathrm{GL}}_n(q),$$

which is called an affine group.

Remark. When $n=1$, ${\mathrm{GL}}_1(q)$ is cyclic and it is isomorphic to ${\mathbb{F}}_q^{\times }$. Assume that $q=p^f$. Recall that a field $F={\mathbb{F}}_{p^f}$ can be viewed as a vector space $F^{+}={\mathbb{F}}_p^f=V$, of dimension $f$ over ${\mathbb{F}}_p$. Then ${\mathrm{AGL}}_1(q)\cong {\mathbb{F}}_q^{+}:{\mathbb{F}}_q^{\times }$ acts on $V$ as linear transformations:

$$(a,b):x\mapsto a+xb,\forall x\in {\mathbb{F}}_p^f.$$

Theorem

Let $G=N:H⩽{\mathrm{AGL}}_1(p^n)$ with $N\cong {\mathbb{Z}}_p^n$, and let $H⩽{\mathbb{Z}}_{p^n-1}$ with $|H|\nmid p^f-1$ for any $f\mid n,f<n$. Then $\mathrm{Aut}(G)\cong \mathrm{A}{\mathrm{\Gamma L}}_1(p^n)={\mathrm{AGL}}_1(p^n):⟨\varphi ⟩$, where $\varphi$ is the Frobenius automorphism of order $n$.

\begin{proof} For any $\sigma \in \mathrm{Aut}(G)$, there is $N^{\sigma }/(N^{\sigma }\cap N)⊲H$ as $N^{\sigma }⊲G$. It yields that $N^{\sigma }=N$ by $(|N|,|H|)=1$. Therefore, $N$ is a characteristic subgroup of $G$. In addition, since $Z(G)$ is trivial, $G\cong \mathrm{Inn}(G)⊲\mathrm{Aut}(G)$. Therefore, we have that $N⊲\mathrm{Aut}(G)$.

Let $S=\{H^{\sigma }:\sigma \in \mathrm{Aut}(G)\}$. Since all complements of $N$ are conjugate in $G$ by Hall’s theorem, $N$ is transitive on $S$. As $\mathrm{Aut}(G)$ is transitive on $S$, by Frattini’s argument we have that $\mathrm{Aut}(G)=N{G}_{(S)}$, where $G_{(S)}$ is the kernel of $\mathrm{Aut}(G)$ acting on $S$. If there exists a nontrivial $n\in N$ such that $H^n=H$, then for any $b\in H$, there is $b^n\in H$. Note that $b^n\in {\mathrm{AGL}}_1(p^n)$ maps $x$ to $bx+(nb-b)$, so $nb=b$ for all $b\in H$. Thus $n=0$, contradiction. Therefore, the action of $N$ on $S$ is regular. So $G_{(S)}\cap N$ is trivial and $\mathrm{Aut}(G)=N:L$ where $L=G_{(S)}$. Since $G⊲\mathrm{Aut}(G)$, $H$ is a normal subgroup of $L$.

Let $K=\{\alpha \in L:h^{\alpha }=h,\forall h\in H\}$. Then $H⩽K⊲L$ and $L$ can be written as $L=K.L_0$. For any $\alpha \in K$, there is $v^{\alpha h}=v^{h\alpha }$ for all $v\in N$. Since $N$ can be seen as ${\mathbb{F}}_{p}H$-module, $\alpha$ is a module isomorphism of $N$. Note that $N$ is irreducible. Thus by Schur’s lemma, $K$ is a multiplicative group of a field. Since $K⩽\mathrm{Aut}(N)\cong {\mathbb{Z}}_{p^n-1}$ and $|H|\nmid p^f-1$ for all $f\mid n$ with $f<n$, we obtain that $K\cong {\mathbb{Z}}_{p^n-1}$ by this. Hence, $\mathrm{Aut}(G)\cong ({\mathbb{Z}}_p^n:{\mathbb{Z}}_{p^n-1}).\widetilde{L_0}$ with $\widetilde{L_0}\cong L_0$ and so $\mathrm{Aut}(G)⩽\mathrm{Aut}({\mathrm{AGL}}_1(p^n))$.

We claim that $\mathrm{Aut}({\mathrm{AGL}}_1(p^n))=\mathrm{A}{\mathrm{\Gamma L}}_1(p^n)$. Let $A:=\mathrm{Aut}({\mathrm{AGL}}_1(p^n))$, and let ${\mathrm{AGL}}_1(p^n)=F^{+}:F^{\times }$ where $F={\mathbb{F}}_{p^n}$. Similarly, by Hall’s theorem each complement of $F^{+}\cong {\mathbb{Z}}_p^n$ in ${\mathrm{AGL}}_1(p^n)$ is in the same conjugacy class, and so there exists $a\in F^{+}$ such that $(F^{\times }{)}^{\delta a}=F^{\times }$. Then we may assume that $(F^{\times }{)}^{\delta }=F^{\times }$ and $\delta (1)=1$ as $a\in \mathrm{Inn}({\mathrm{AGL}}_1(p^n))$.

For $0\ne a\in F^{+}$, use $\bar{a}$ to denote the corresponding element of $F^{\times }$. So the semidirect product action is $\bar{a}b{\bar{a}}^{-1}=ab$. Now, for $0\ne a\in F^{\times }$, using $\delta (1)=1$, we have

$$\delta (a)=\delta \left(\bar{a}1{\bar{a}}^{-1}\right)=\delta (\bar{a})1\delta (\bar{a}{)}^{-1},$$

so $\overline{\delta (a)}=\delta (\bar{a})$. In other words $\delta$ is acting in the same ways on $F^{+}\setminus \{0\}$ and on $F^{\times }$. Then, for $a,b\in F^{+}\backslash \{0\}$

$$\delta (a)\delta (b)=\overline{\delta (a)}\delta (b){\overline{\delta (a)}}^{-1}=\delta (\bar{a})\delta (b)\delta (\bar{a}{)}^{-1}=\delta \left(\bar{a}b{\bar{a}}^{-1}\right)=\delta (ab)$$

and hence $\delta$ is acting as a field automorphism of $F$. Therefore, $\mathrm{Aut}({\mathrm{AGL}}_1(p^n))=\mathrm{A}{\mathrm{\Gamma L}}_1(p^n)$.

Furthermore, since $\mathrm{A}{\mathrm{\Gamma L}}_1(p^n)$ acts on $G=N:H$ faithfully, we obtain that $\mathrm{Aut}(G)=\mathrm{A}{\mathrm{\Gamma L}}_1(p^n)$. \end{proof}