Exercise 12.1.21

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}}

Part (a) of Theorem 12.1.10 implies that when $n\geq 5$, the index of a proper subgroup of $S_n$ is either $2$ or $\geq n$.
\be
\item[(a)] Prove that $S_n$ always has a subgroup $H$ of index $n$. This means that equality can occur in the bound $[S_n:H] \geq n$.

\item[(b)] Give an example to prove that Theorem 12.1.10 is false when $n=4$.
\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}
\be
\item[(a)] The subgroup $H$ of $S_n$ of the permutations $\sigma$ that fix $n$ is a subgroup isomorphic to $S_{n-1}$, and $[S_n : H ] = n!/(n-1)! = n$.
\item[(b)] In the Exercise 3, we saw that $H = H(y_1)$, where $y_1 = x_1x_2+x_3x_4$ is a group isomorphic to $D_8$:
$$\langle (1\, 2), (1\,3\,2\,4) angle= \{(), (1\,2), (1\,3\,2\,4), (1\,3)(2\,4), (1\,2)(3\,4), (1\,4)(2\,3), (3\,4), (1\,4\,2\,3)\},$$
so $[S_4 : H] = 3 <n=4$. This proves that the Theorem 12.1.10 is false if we forget the hypothesis $n\geq 5$.
\ee
\end{proof}

Exercise 12.1.21

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Exercise 12.1.21

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