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Let $\mathbb{F_q}$ be a finite field. Consider the group Aff$\mathbb{(F_q)}$

Aff$\mathbb{(F_q)} := $ $ \ \begin{Bmatrix} \begin{pmatrix} a&b\\ 0&1\\ \end{pmatrix} \colon a, b \in \mathbb{F_q}, a \neq 0 \end{Bmatrix} $.

Let $H := $ $ \ \begin{Bmatrix} \begin{pmatrix} 1&b\\ 0&1\\ \end{pmatrix} \colon b \in \mathbb{F_q} \end{Bmatrix} $ be a normal subgroup of Aff$\mathbb{(F_q)}$, which is isomorphic to $\mathbb{F_q}$ as groups.

How to describe all extension groups of $H$ by Aff$\mathbb{(F_q)}/H$?

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  • $\begingroup$ We have $Aff(q)/H\cong \mathbb{F}_q^*$, so that the equivalence classes of extensions are described by $H^2(\mathbb{F}_q^*,\mathbb{F}_q)$. $\endgroup$ – Dietrich Burde May 29 '16 at 15:34
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By the Schur-Zassenhaus theorem, every extension $$ 1\rightarrow M\rightarrow E\rightarrow G\rightarrow 1 $$ with $gcd(|M|,|G|)=1$ is split, i.e., $E\cong M\rtimes G$. In other words, we have $H^2(G,M)=1$.

We can apply this here with $G=Aff(q)/H\cong \mathbb{F}_q^*$ of order $q-1$ and $M=H\cong \mathbb{F}_q$ of order $q$, because $gcd(q,q-1)=1$. So there are only split extensions, i.e., semidirect products.

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  • $\begingroup$ Your second ($\mathrm H^2(G, M) = 1$) and last ("there is only the split extension"—my emph) sentences seem to be misleading; there can be more than one split extension. In fact, there is one for every divisor $d$ of $q - 1$, where the image of $m \in M = \mathbb F_q$ under the action of $g \in G = \mathbb F_q^\times$ is $g^d m$. (Extensions coming from integers $d$ that don't divide $q - 1$ are isomorphic to those coming from $\mathrm{gcd}(d, q - 1)$.) Note, for example, that the $d = q - 1$ extension (direct product) is Abelian, and the $d = 1$ one (natural semi-direct product) isn't. $\endgroup$ – LSpice Sep 4 '16 at 14:37

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