Exercise 14.3.3

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Consider the affine semilinear group $\mathrm{A\Gamma L}(n,\F_q)$ for $q = p^m$.
 \be
 \item[(a)] Prove that $\mathrm{AGL}(n,\F_q)$ is a normal subgroup of $\mathrm{A\Gamma L}(n,\F_q)$ of index $m$.
 \item[(b)] Prove that $\F_q^n$ is a normal subgroup of $\mathrm{A\Gamma L}(n,\F_q)$.
 \item[(c)] Prove that elements of $\mathrm{A\Gamma L}(n,\F_q)$ give maps $\F_q^n 	o \F_q^n$ that are affine linear over $\F_p$.
 \ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\bigskip
 
 We give some preliminary formulas.
 
Note that $\gamma_{A,v} = \gamma_{A,e,v}, \ A \in \mathrm{GL}(n,\F_q), v \in \F_q$, where $e$ is the identity element of $\Gal(F_{q^m}/\F_p)$, thus $\mathrm{AGL}(n,\F_q) \subset \mathrm{A\Gamma L}(n,\F_q)$.

For all $u = (\alpha_1,\ldots,\alpha_n) \in \F_q^n$, we write $\sigma(u) = (\sigma(\alpha_1),\ldots,\sigma(\alpha_n))$.

If $A =(a_{ij})_{1\leq i,j \leq n} \in \mathrm{GL}(n,\F_q)$, then 
\begin{align*}
\sigma(A\cdot u) &= \sigma\left(\sum_{j=1}^n a_{1j} \alpha_j,\ldots,\sum_{j=i}^n a_{ij} \alpha_j,\ldots,\sum_{j=1}^n a_{nj} \alpha_jight)\
&=\left(\sum_{j=1}^n \sigma(a_{1j}) \sigma(\alpha_j),\ldots,\sum_{j=i}^n \sigma(a_{ij})\sigma( \alpha_j),\ldots,\sum_{j=1}^n \sigma(a_{nj})\sigma( \alpha_j)ight)\
&=\sigma(A)\cdot \sigma(u),
\end{align*}
where $\sigma(A) = (\sigma(a_{ij}))_{1\leq i,j \leq n}$, and similarly 
$$\sigma(A\cdot u + v) = \sigma(A)\cdot  \sigma(u) + \sigma(v).$$
Then, for all $u \in \F_q^n$,
\begin{align*}
(\gamma_{A,\sigma,v} \circ \gamma_{B,	au,w})(u) &= A\sigma(B	au(u)+w) + v\
&=A\sigma(B) (\sigma	au)(u) + A\sigma(w) + v\
&=\gamma_{A\sigma(B),\sigma 	au,A \sigma(w) + v}(u),
\end{align*}
thus
\begin{align}
\gamma_{A,\sigma,v} \circ \gamma_{B,	au,w} = \gamma_{A\sigma(B),\sigma 	au,A \sigma(w) + v}.
\end{align}

For all $u,s \in \F_q^n$,
\begin{align*}
s = \gamma_{A,\sigma,v}(u) &\iff s = A \sigma(u) + v\
&\iff \sigma(u) = A^{-1}(s -v)\
&\iff u = \sigma^{-1}(A^{-1} s - A^{-1}v)\
&\iff u = \sigma^{-1}(A^{-1}) \sigma^{-1}(s) - \sigma^{-1}(A^{-1})\sigma^{-1}(v)\
&\iff u = \gamma_{\sigma^{-1}(A^{-1}),\sigma^{-1},- \sigma^{-1}(A^{-1})\sigma^{-1}(v)}(s),
\end{align*}
thus
\begin{align}
\gamma_{A,\sigma,v}^{-1} = \gamma_{\sigma^{-1}(A^{-1}),\sigma^{-1},- \sigma^{-1}(A^{-1})\sigma^{-1}(v)}.
\end{align}
By (6) and (7), $\mathrm{A\Gamma L}(n,\F_q)$ is a subgroup of $S(\F_q^n)$.
Moreover, if $\gamma_{A,\sigma,v}, \gamma_{B,	au,w} \in  \mathrm{A\Gamma L}(n,\F_q)$, then
\begin{align}
 \gamma_{A,\sigma,v} =  \gamma_{B,	au,w} \Rightarrow (A,\sigma,v) = (B,	au,w).
\end{align}
 Indeed, if $\gamma_{A,\sigma,v} =  \gamma_{B,	au,w}$, then for all $u \in \F_q^n$, $A \sigma(u) + v = B 	au(u) + w$. With $u=0$, we obtain $v = w$. If we take $u =e_i =(0,\ldots,0,1,0,\ldots,0)$, where $(e_1,\ldots,e_n)$ is the standard base of $\F_q^n$, then $\sigma(e_i) = e_i = 	au(e_i)$, so that $Ae_i = Be_i,\ i=1,\ldots,n$. This implies $A = B$, and $\sigma(u) = 	au(u)$ for all $u\in \F_q^n$. Therefore, with $u = (\alpha,0,\ldots,0),\ \alpha \in \F_q$, we obtain $\sigma(\alpha) = 	au(\alpha)$, thus $\sigma = 	au$.

\begin{proof}
 \item[(a)] 
 By (8),
 $$
 \varphi
 \left\{
 \begin{array}{ccc}
 \mathrm{A\Gamma L}(n,\F_q) &	o & \Gal(\F_{p^m},\F_p)\
 \gamma_{A,\sigma,v} & \mapsto & \sigma
 \end{array}
 ight.
 $$
 is well defined. Moreover, by (6), if $\gamma_{A,\sigma,v}, \gamma_{B,	au,w} \in \mathrm{A\Gamma L}(n,\F_q)$,
 $$\varphi(\gamma_{A,\sigma,v} \circ \gamma_{B,	au,w} ) = \varphi( \gamma_{A\sigma(B),\sigma 	au,A \sigma(w) + v}) =\sigma 	au = \varphi(\gamma_{A,\sigma,v})\varphi( \gamma_{B,	au,w}),
 $$
 thus $\varphi$ is a group homomorphism. If $\sigma$ is any element of $\Gal(\F_{p^n},\F_p)$, then $\sigma = \varphi(\gamma_{I_n, \sigma,0})$, thus $\varphi$ is surjective. The kernel of $\varphi$ is
 $$\ker \varphi = \{\gamma_{A,e,v}\, | \, A\in \mathrm{GL}(n,\F_q), v \in \F_q^n\} = \{\gamma_{A,v}\, | \, A\in \mathrm{GL}(n,\F_q), v \in \F_q^n\} = \mathrm{AGL}(n,\F_q).$$

This shows that $\mathrm{AGL}(n,\F_q)$ is a normal subgroup of $\mathrm{A\Gamma L}(n,\F_q)$, and the first Isomorphism Theorem gives
$$ \mathrm{A\Gamma L}(n,\F_q)/ \mathrm{AGL}(n,\F_q) \simeq \Gal(\F_{p^m},\F_p).$$
Since $\Gal(\F_{p^m},\F_p)$ is a cyclic group of order $m$, $\mathrm{AGL}(n,\F_q)$ is a normal subgroup of $\mathrm{A\Gamma L}(n,\F_q)$ of index $m$.

\item[(b)] Here, $\F_q^n$ is identified with the group of translations $\{\gamma_{I_n,e,w}\,|\, w \in \F_q^n\}$, and, using (6) and (7),
\begin{align*}
\gamma_{A,\sigma,v} \circ \gamma_{I_n,e,w} \circ \gamma_{A,\sigma,v}^{-1} &= \gamma_{A,\sigma,v} \circ \gamma_{I_n,e,w}  \circ \gamma_{\sigma^{-1}(A^{-1}),\sigma^{-1},- \sigma^{-1}(A^{-1})\sigma^{-1}(v)}\
&= \gamma_{A,\sigma,v} \circ  \gamma_{\sigma^{-1}(A^{-1}),\sigma^{-1},-\sigma^{-1}(A^{-1}) \sigma^{-1}(v) + w}\
&=\gamma_{A \sigma \sigma^{-1}(A^{-1}),\sigma \sigma^{-1}, A\sigma(-\sigma^{-1}(A^{-1}) \sigma^{-1}(v) + w) + v}\
&= \gamma_{I_n,e,A\sigma(w)}.
\end{align*}
We have proved
\begin{align}
\gamma_{A,\sigma,v} \circ \gamma_{I_n,e,w} \circ \gamma_{A,\sigma,v}^{-1} = \gamma_{I_n,e,A\sigma(w)} \in \F_q^n,
\end{align}
therefore $\F_q^n$ is a normal subgroup of $\mathrm{A\Gamma L}(n,\F_q)$.

\item[(c)]  $\F_q^n$, which is a vector space over $\F_q$, is also a vector space over $\F_p$, by restriction of the extern operation 
$
\left\{
\begin{array}{ccc}
\F_q 	imes \F_q^n & 	o & \F_q^n\
(\lambda,u) & \mapsto &\lambda u
\end{array}
ight.
$
to $\F_p 	imes \F_q^n$.

Let ${\cal B} = (e_1,\ldots,e_m)$ be a base of $\F_q$ over $\F_p$.
As $\F_p$-vector spaces, $\F_q^n$ and $\F_p^{nm}$ are isomorphic, where an isomorphism $\varphi$ is given by
$$
\varphi \left\{
\begin{array}{ccc}
\F_q^n & 	o & \F_p^{nm}\
(\alpha_1,\ldots,\alpha_n)& \mapsto &(x_1^1,\ldots,x_m^1;\ldots;x_1^n,\ldots,x_m^n),
\end{array}
ight.
$$
where $\alpha_i = \sum_{j=1}^m x_j^i e_j,\ i=1,\ldots,n.$ (this isomorphism depends of the choice of the base $\cal B$). This proves $\dim_{\F_p} \F_q^n = nm$.
Consider the two maps
$$
\psi
\left\{
\begin{array}{ccc}
\F_q^n &	o &\F_q^n\
u  = (\alpha_1,\ldots,\alpha_n)& \mapsto &\sigma(u) = (\sigma(\alpha_1),\ldots,\sigma(\alpha_n)),
\end{array}
ight.
\qquad 
\chi
\left\{
\begin{array}{ccc}
\F_q^n &	o &\F_q^n\
v& \mapsto &A \cdot v.
\end{array}
ight.
$$
The automorphism $\sigma \in \Gal(\F_q/\F_p)$ is $\F_q$-linear: if $\lambda \in \F_p, \alpha \in \F_q$, $\sigma(\lambda \alpha) = \sigma(\lambda) \sigma(\alpha) = \lambda \sigma(\alpha)$. Thus $\psi$ is $\F_p$-linear. Moreover, $\chi$ is $\F_q$-linear, a fortiori $\F_p$-linear. Therefore, $\chi \circ \psi = u \mapsto A\cdot \sigma(u)$ is $\F_p$-linear, so that
$$
\gamma_{A,\sigma,v}
\left\{
\begin{array}{ccc}
\F_q^n & 	o & \F_q^n\
u &\mapsto & A\cdot \sigma(u) + v
\end{array}
ight.
$$
is affine linear over $\F_p$.
 \end{proof}

Exercise 14.3.3

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Exercise 14.3.3

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