I am wondering what the structure of the automorphism group of the general affine group of the affine line over a finite field looks like. I'll make that a bit more precise:

If $k$ is a finite field, and $\operatorname{AGL}_1(k)$ its group of affine transformations, i.e. maps of the form $$k\ \longrightarrow\ k:\ x\ \longmapsto\ ax+b,$$ with $a\in k^{\times}$ and $b\in k$, then what is the isomorphism type of $\operatorname{Aut}(\operatorname{AGL}_1(k))$?

I know that $\operatorname{AGL}_1(k)\cong k\rtimes k^{\times}$, where the semi-direct product is given by the natural action of $k^{\times}$ on $k$ by multiplication. Also, as the center of $\operatorname{AGL}_1(k)$ is trivial, it is isomorphic to a subgroup of its isomorphism group. Any automorphism of $\operatorname{AGL}_1(k)$ restricts to a group automorphism of $k^{+}$, of which there are very many, unfortunately.

What is a good way to approach this problem?

  • $\begingroup$ Most folks would assume that you wanted to consider only those maps that come from the identity automorphism of $k$. Are you sure you want that full generality? $\endgroup$
    – Lubin
    Feb 19, 2015 at 4:09
  • $\begingroup$ What do you mean by 'come from' the identity automorphism of $k$? I would like to understand the structure of the entire automorphism group, so yes I would like that full generality. $\endgroup$
    – user38292
    Feb 19, 2015 at 4:12
  • $\begingroup$ Then before all else, you have to decide on what the automorphism group of $k^+$ is. It’s a finite vector space over the prime field $\Bbb F_p$, so you’re talking about a general linear group over the prime field. These are well understood… $\endgroup$
    – Lubin
    Feb 19, 2015 at 4:15
  • $\begingroup$ I understand that the automorphism group of $\Bbb{F}_{p^k}$ is (isomorphic to) $\operatorname{GL}_k(\Bbb{F}_p)$. But I think not all automorphisms of a finite field can 'extend' to automorphisms of the affine group? $\endgroup$
    – user38292
    Feb 19, 2015 at 4:17
  • $\begingroup$ Why not? Do you have an example of an unextendable one? $\endgroup$
    – Lubin
    Feb 19, 2015 at 4:18

1 Answer 1


Let $G = {\rm AGL}_1(k)$. Then ${\rm Aut}(G) = {\rm A \Gamma L}_1(k) \cong G\langle \gamma \rangle$, where $\gamma$ is a generator of the group of field automorphisms of $k$. So if $|k|=p^e$ with $p$ prime, then $|\gamma|=e$ and you $\gamma:x \mapsto x^p$ for $x \in k$.

Here is a sketch proof. Any $\alpha \in {\rm Aut}(G)$ must fix the normal subgroup $k$ of the semidirect product and, since the complements are all conjugate in $G$, we can assume (by multiplying $\alpha$ by an inner automorphism) that if fixes the principal complement $k^\times$. Since $k^\times$ acts transitively by conjugation on $k \setminus \{0\}$, by multiplying $\alpha$ by an inner automorphism again, we can assume that $\alpha(1)=1$.

For $0 \ne a \in k < G$, I will use $\bar{a}$ to denote the corresponding element of the complement $k^\times$. So the semidirect product action is $\bar{a}b\bar{a}^{-1} = ab$.

Now, for $0 \ne a \in K$, using $\alpha(1)=1$, we have $$\alpha(a) = \alpha(\bar{a}1\bar{a}^{-1}) =\alpha(\bar{a})1\alpha(\bar{a})^{-1},$$ so $\overline{\alpha(a)} = \alpha(\bar{a})$. In other words $\alpha$ is acting in the same ways on $k -\{0\}$ and on $k^\times$.

So, for $a,b \in k \setminus \{0\}$, $$\alpha(a)\alpha(b) = \overline{\alpha(a)} \alpha(b) \overline{\alpha(a)} ^{-1} = \alpha(\bar{a}) \alpha(b)\alpha(\bar{a})^{-1} = \alpha(\bar{a}b\bar{a}^{-1}) = \alpha(ab)$$ and hence $\alpha$ is acting as a field automorphism of $k$.

  • $\begingroup$ Ah, that’s good. This looks more like the situation with collineations of the (desarguesian) projective plane. Thank you. $\endgroup$
    – Lubin
    Feb 19, 2015 at 13:55
  • $\begingroup$ Could you explain a little bit more "since complements are all conjugate in $G$, we can assume (by multiplying $\alpha$ by an inner automorphism) that it fixes the principal complement"? We can think $k^{\times} = \{(0,a): a \in k^{\times}\}$ and if we apply conjugation on $k^{\times}$ by some element $(b,c)$ in the semidirect product, then $(b,c) (0,a)(b,c)^{-1} = (b,ca)(-bc^{-1},c^{-1}) = (b-ba,a)$. Thus $k^{\times}$ seems not fixed by conjugation. $\endgroup$
    – user124697
    Jun 20, 2019 at 15:56
  • 1
    $\begingroup$ @user124697 The complements have order $|k|-1$, coprime to $|k|$, so they are all conjugate by the Schur-Zassenhaus Theorem. So if $\alpha$ maps $k^\times$ to $C$, say, then $C$ is conjuagte to $k^\times$ by some element $g \in G$. Now if we compose $\alpha$ with the inner automorphism of $G$ induced by $g^{-1}$, the resulting automorphism fixes $k^\times$. $\endgroup$
    – Derek Holt
    Jun 20, 2019 at 21:28
  • $\begingroup$ @DerekHolt Do you not need that the resulting automorphism is the identity on $k^{\times}$ for the argument to go through? It could be that after conjugation, the result is some non-trivial automorphism of $k^{\times}$. I think this can't actually happen, but it takes some effort/computation to show that. $\endgroup$ Nov 30, 2021 at 6:05
  • $\begingroup$ @MatthewDrury Sorry but I cannot make any sense of your question. I never said anything about the resulting automorphism being the identity on $k^\times$. If it was then the automorphism would be the identity map. $\endgroup$
    – Derek Holt
    Nov 30, 2021 at 8:09

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