Exercise 14.3.10

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}}

A permutation group $G \subset S_l$ is regular if there is a  one-to-one onto map $\gamma : G 	o \{1,\ldots,l\}$ such that $\hat \gamma : S(G) \simeq S_l$ maps $\{\varphi_g \mid g \in G\} \subset S(G)$ to $G \subset S_l$. Recall that $\varphi_g \in S(G)$ is defined by $\varphi_g(h) = gh$ for $h\in G$. The goal of this exercise is to show that $G$ is regular if and only if it is transitive with trivial isotropy subgroups.
  \be
  \item[(a)] Let $G \subset S_l$ be regular. Prove that $G$ is transitive and that the isotropy subgroups of $G$ are trivial.
  \item[(b)] For the rest of the exercise, assume that $G$ is transitive with trivial isotropy subgroups. Define $\gamma : G 	o \{1,\ldots,l\}$ by $\gamma(	au) = 	au(1)$ for $	au \in G$. Prove that this map is one-to-one and onto.
  \item[(c)] The map $\gamma$ of part (b) gives $\hat \gamma:S(G) \simeq S_l$. Show that $\hat \gamma(\varphi_g) = g$, and conclude that $G$ is regular.
  \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)] Recall that $\hat \gamma(\sigma) = \gamma \circ \sigma \circ \gamma^{-1}$ for all $\sigma \in S_l$.
  
  Let $i,j$ be any elements in $\{1,\ldots,l\}$. We must find $g' \in G$ such that $g'(i) = j$.
  
  Take $g = \gamma^{-1}(j) [\gamma^{-1}(i)]^{-1} \in G$, and $g' = \hat \gamma(\varphi_g)$. Then $g' \in G$ since $\hat \gamma$ maps ${\{\varphi_g \mid g \in G\}}$ to $G$.
  
  Then $g \gamma^{-1}(i) = \gamma^{-1}(j)$, thus $(\gamma \circ \varphi_g \circ \gamma^{-1})(i) = j$, that is $g' (i) = (\hat \gamma(\varphi_g))(i) = j$, where $g' \in G$. This proves that $G$ is transitive.
  
  Let $i$ be any element in $\{1,\ldots,l\}$, and take  $g' \in G_i$, where $G_i$ is the isotropy subgroup of $i$, so that $g'(i) = i$. Since $\hat \gamma$ maps $\{\varphi_g \mid g \in G\} \subset S(G)$ to $G$, there exists some $g \in G$ such that $g' = \hat \gamma(\varphi_g) =\gamma \circ \phi_g \circ \gamma^{-1}$.
  
  Then $g'(i) = i$ implies $g \gamma^{-1}(i) = \gamma^{-1}(i)$, thus $g = e , \phi_g = e$, and $g' = \gamma \circ \phi_g \circ \gamma^{-1} = e$, where $e = 1_G$ is the identity element of $G$. Therefore $G_i=\{e\}$ for all $i \in \{1,\ldots,l\}$.
  
 \item[(b)] We prove first that $\gamma$ is injective. Suppose that $\gamma(	au) = \gamma(	au')$, where $	au,	au' \in G$. Then $	au(1) = 	au'(1)$, thus $(	au^{-1} \circ 	au')(1) = 1$, and $	au^{-1} \circ 	au' \in G_1$, where $G_1$ is the isotropy subgroup of $1$. But $G_1 = \{e\}$, thus $	au^{-1} \circ 	au' = e$, and $	au = 	au'$. We have proved that $\gamma$ is injective.
 
 Now take $i$ be any element in $\{1,\ldots,l\}$. Since $G$ is transitive, there is some $	au \in G$ such that $	au(1) = i$, that is $\gamma(	au) = i, 	au \in G$. This proves that $\gamma$ is surjective.
 
 $\gamma : G 	o \{1,\ldots,n\}$ is bijective.
 
 \item[(c)] let $i$ be any element in $\{1,\ldots,l\}$. Since $G$ is transitive, there is some $	au \in G$ such that $	au(1) = i$, thus $\gamma(	au) = i$. Therefore
 \begin{align*}
 [\hat \gamma(\varphi_g)](i) &= (\gamma \circ \varphi_g \circ \gamma^{-1})(i)\
 &= (\gamma \circ \varphi_g)(	au)\
 &= \gamma(g	au)\
 &= (g	au)(1) = g(i).
 \end{align*}
 Since this is true for all $i$,
 $$\hat \gamma(\varphi_g) = g.$$

This proves that the isomorphism $\hat \gamma$ maps $\{\varphi \mid g \in G\}$ in $G$, and since these two subgroup have same order $l$, $\hat \gamma$ maps $\{\varphi_g\mid g \in G\}$ on $G$.

$G$ is a regular subgroup of $S_l$.

\end{proof}

Note: This proves also that $G \subset S_l$ is regular iff there is some bijective $\gamma : G 	o \{1,\ldots l\}$ such that $\hat \gamma$ maps $\varphi_g$ to $g$, for every $g\in G$.

Exercise 14.3.10

Answers

Comments

Exercise 14.3.10

Answers

Comments

Add answer