Exercise 14.3.4

Question

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

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ewcommand{\ssi} {si et seulement si }

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

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

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

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

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

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

Let $F$ be any field. The definition of $\mathrm{AGL}(n,\F_q)$ given in the text extend to $\mathrm{AGL}(n,F)$.
The goal of this exercise is to prove that $\mathrm{AGL}(n,F)$ is doubly transitive when we regard elements of $\mathrm{AGL}(n,F)$ as permutations of the vector space $F^n$.
\be
\item[(a)] Use $F^n \subset \mathrm{AGL}(n,F)$ to show that $\mathrm{AGL}(n,F)$ acts transitively on $F_n$.
\item[(b)] Inside $\mathrm{AGL}(n,F)$, we have the isotropy subgroup of $\,0\in F_n$. Prove that this isotropy subgroup is $\mathrm{GL}(n,F)$.
\item[(c)] Prove that $\mathrm{GL}(n,F)$ acts transitively on $F^n\setminus \{0\}$.
\item[(d)]Use Exercise 19 below to conclude that $\mathrm{AGL}(n,F)$ is doubly transitive.
\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}}

\begin{proof}
\item[(a)] Let $u,u'$ be any vectors in $F^n$. The equality $I_n\cdot u  + (u'- u) = u'$ shows that $\gamma_{I_n,u'-u}(u) = u'$, where $\gamma_{I_n,u'-u} \in \mathrm{AGL}(n,F)$.

Therefore $\mathrm{AGL}(n,F)$ acts transitively on $F^n$.

\item[(b)] Write $G_0$ the isotropy group of $0$. Then
$$\gamma_{A,v} \in G_0 \iff A\cdot 0 + v = 0 \iff v = 0.$$
Therefore $G_0 = \{\gamma_{A,0}\, | \,  A \in \mathrm{GL}(n,F)\} \simeq \mathrm{GL}(n,F)$.

In section 14.3.B, we identified $\{\gamma_{A,0}\mid A \in \mathrm{GL}(n,F)\} $ with $\mathrm{GL}(n,F)$, so that
$$G_0 = \mathrm{GL}(n,F).$$

\item[(c)]

Let $u,v \in F^n \setminus \{0\}$. Since $u 
e 0$, we can complete $u$ in a base ${\cal B} _1= (u_1,\ldots,u_n)$ of $F_n$, where $u_1 = u$. Similarly, we can complete $v
e 0$ in a base ${\cal B}_2 = (v_1,\ldots,v_n)$, where $v_1 = v$.

Since ${\cal B}_1,{\cal B}_2$ are two bases, there exists some bijective linear map $f : F^n 	o F^n$ such that $f(u_i) = v_i,\ i=1,\ldots,n$, so that $f(u) = v$.

Let ${\cal B} = (e_1,\ldots,e_n)$ be the standard base of $F^n$. Then the matrix $A = {\cal M}_{\cal B}(f)$ of $f$ satisfies $A \in \mathrm{GL}(n,F)$, and $A \cdot u = v$. This proves that $\mathrm{GL}(n,F)$ acts transitively on $F^n \setminus\{0\}$.

\item[(d)] Since  the isotropy group $G_0$ acts transitively on $F^n\setminus \{0\}$, Exercise 14.3.19 (a) shows that $\mathrm{AGL}(n,F)$ acts transitively on $F^n$, using that  $\mathrm{AGL}(n,F)$ can be viewed as a subgroup $G$ of $S_{|F^n|}$.
\end{proof}

Exercise 14.3.4

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Exercise 14.3.4

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