Exercise 14.4.19

Question

\def\imp{\Rightarrow}
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ewcommand{\ssi} {si et seulement si }

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

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

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

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

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

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

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

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In the Mathematical Notes, we mentioned that all subgroups of $\mathrm{PSL}(2,\F_q)$ are known up to conjugacy. We will do a small part of this classification by proving that $\mathrm{PSL}(2,\F_q)$ contains a subgroup isomorphic to $S_4$ when $p\equiv 1 \pmod 8$. To begin, note that by Exercise 10, the images of the matrices (14.25) generate a subgroup of $\mathrm{PGL}(2,\F_q)$ isomorphic to $S_4$.
\be
\item[(a)] Explain why $\F_p^*$ has an element $\zeta$ of order $8$. Then $i = \zeta^2$ has order $4$.
\item[(b)] Compute $(1+i)^2$ and use this to prove that there is $\alpha \in \F_p$ such that $\alpha^2 = 2$.
\item[(c)] Show that the matrices (14.25) lie in $\mathrm{SL}(2,\F_p)$ after multiplication by suitable elements of $\F_p^*$. Hence their images generate a subgroup of $\mathrm{PSL}(2,\F_p)$ isomorphic to $S_4$.
\item[(d)] Over $\C$, $\zeta_8 = \cos(2\pi/8) + i \sin(2\pi/8) = (1+i)/\sqrt{2}$. How does this relate to part (b)?
\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)] We know that $\F_p^*$ is cyclic. Let $g$ a generator of $\F_p^*$. Then the order of $g$ is $p-1$, and $8 \mid p-1$, thus $\zeta = g^{(p-1)/8}$ has order $8$ (and $i = \zeta^2$ has order $4$, therefore $i^2 = -1$).
\item[(b)] Since $i^2 = -1$, $(1+i)^2 = 2i = 2 \zeta^2$, thus $\left(\frac{1+i}{\zeta}ight)^2 = 2$. If 
$$\alpha = \frac{1+i}{\zeta} = \frac{1+\zeta^2}{\zeta} = \zeta + \zeta^{-1},$$ then $\alpha \in \F_p$, and $\alpha^2 = 2$.
\item[(c)] Consider the matrices (14.25):
$$g = \begin{pmatrix} 0 & 1\1 & 0\end{pmatrix},\qquad  h = \begin{pmatrix} i & 0\0 & 1 \end{pmatrix},\qquad   k= \begin{pmatrix} 1 & -1\1 & 1 \end{pmatrix}.$$
In Exercise 10, we have proved that, if $p\equiv 1 \pmod 8$,
$$\langle [g],[h],[k]angle = \left \langle
\begin{bmatrix} 0 & 1\1 & 0\end{bmatrix}, \begin{bmatrix} i & 0\0 & 1 \end{bmatrix},  \begin{bmatrix} 1 & -1\1 & 1 \end{bmatrix}
 ight angle = N(H) \simeq S_4,
$$
where $N(H) \subset \mathrm{PGL}(2,\F_p)$.
Note that 
$$\det(f) = -1 = i^2,\qquad \det(h) = i = \zeta^2,\qquad \det(k) = 2 = \alpha^2, $$
where $i,\zeta, \alpha \in \F_p^*$, so that
$$\frac{1}{i} g \in \mathrm{SL}(2,\F_p), \qquad \frac{1}{\zeta} h  \in \mathrm{SL}(2,\F_p), \qquad \frac{1}{\alpha} k  \in \mathrm{SL}(2,\F_p).$$
Thus $[g] = [\frac{1}{i} g], [h] = [ \frac{1}{\zeta} h], [k] = [ \frac{1}{\alpha} k]$ are in $\mathrm{PSL}(2,\F_p)$. Therefore
$$S_4 \simeq N(H) = \langle [g],[h],[k]angle \subset \mathrm{PSL}(2,\F_p).$$
$\mathrm{PSL}(2,\F_p)$ contains a copy of $S_4$ if $p\equiv 1 \pmod 8$.
\item[(d)] We know that $\zeta_8$ is such that 
$$\zeta_8 + \zeta_8^{-1} = \zeta_8 + \overline{\zeta_8} = 2 \cos(\pi/4) = \sqrt{2},$$
so that $(\zeta_8 + \zeta_8^{-1})^2 = 2$ (draw a regular octagon).

If $\zeta$ has order $8$ in $\F_p^*$, as $\zeta_8$ has order $8$ in $\C^*$, it is not surprising that $(\zeta + \zeta^{-1})^2 = 2$,

To give a direct proof in these two groups, note that, since the order of $\zeta$ is $8$,  
$(\zeta^4)^2 = 1$, and $\zeta^4 
e 1$, thus $\zeta^4 = -1 = \zeta^{-4}$. Therefore $(\zeta^2 + \zeta^{-2})^2 = \zeta^4 + \zeta^{-4} + 2 = 0$, so that $\zeta^2 + \zeta^{-2} = 0$. Then 
$$(\zeta + \zeta^{-1})^2 = \zeta^2 + \zeta^{-2} + 2 = 2.$$
This equality is used to compute the quadratic character of $2$ modulo $p$, $\legendre{2}{p} = (-1)^{(p^2-1)/8}$.
\end{proof}

Exercise 14.4.19

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Exercise 14.4.19

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