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

\begin{gathered} \verb|Aff|\mathbb{(F_q)} := \ \begin{Bmatrix} \begin{pmatrix} a&b\\ 0&1\\ \end{pmatrix} \colon a \in \mathbb F^{*}_q, b \in \mathbb{F}_q \end{Bmatrix} \end{gathered}

It's easy to see that the homomorphism $\psi: \mathrm{Aff}(\mathbb{F}_q) \rightarrow \mathbb F^{*}_q$ defined by $\begin{pmatrix} a&b\\ 0&1\\ \end{pmatrix} \mapsto a$ describes some normal subgroups. Every subgroup H of $\mathbb F^{*}_q$ (which is cyclic of order q-1) is normal in $\mathbb F^{*}_q$, so we get a normal subgroup of $\mathrm{Aff}(\mathbb{F}_q)$ by taking the inverse image of H: the group of matrices \begin{Bmatrix} \begin{pmatrix} a&b\\ 0&1\\ \end{pmatrix} \colon a \in H, b \in \mathbb{F_q} \end{Bmatrix} where $H \trianglelefteq \mathbb F^{*}_q \simeq \mathbb{Z_{q-1}}$.

Is that full description?

  • $\begingroup$ Yes, apart from the trivial subgroup, all normal subgroups are of that form. $\endgroup$ – Derek Holt May 29 '17 at 20:49
  • $\begingroup$ @DarekHolt I'm sorry, how to show that there are not other subgroups? $\endgroup$ – user272816 May 29 '17 at 20:54

Note that $K:=\ker \psi = \{ \left( \begin{array}{cc}1&b\\0&1 \end{array}\right):b \in {\mathbb F}_q \}$. Also $$\left( \begin{array}{cc}a&0\\0&1 \end{array}\right)\left( \begin{array}{cc}1&b\\0&1 \end{array}\right)\left( \begin{array}{cc}a&0\\0&1 \end{array}\right)^{-1}=\left(\begin{array}{cc}1&ab\\0&1 \end{array}\right)$$ so all nontrivial elements of $K$ are conjugate in $G := {\rm Aff}({\mathbb F}_q)$. So $[G,G] = K$.

Now let any $N$ be any nontrivial normal subgroup of $G$ and let $g \in N \setminus \{1\}$. Since $Z(G)=1$, there exists $h \in G$ with $[g,h] \ne 1$ and $[g,h] \in [G,G] = K$. Hence $K \le N$, and so $N$ is the inverse image under $\psi$ of a subgroup of ${\mathbb F}_q^*$.

  • $\begingroup$ Sorry, but the product of such matrices is [1 a^{-1}b] [0 1 ] $\endgroup$ – user272816 May 29 '17 at 23:10
  • $\begingroup$ Corrected (although that wouldn't affect the argument). $\endgroup$ – Derek Holt May 30 '17 at 7:34
  • $\begingroup$ @DarekHolt Sorry, but > there exists h \in G with... do you mean there exists h \in N ? $\endgroup$ – user272816 May 30 '17 at 22:19
  • $\begingroup$ No, $g$ was supposed to be in $N$ - i've edited it. $\endgroup$ – Derek Holt May 31 '17 at 8:56
  • $\begingroup$ @DarekHolt I'm sorry, could you explain please, why does [G,G] \leq N? We showed that [N,G] \leq N because we chose g from N. $\endgroup$ – user272816 Jun 26 '17 at 21:16

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