Exercise 14.4.20

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}}

Assume that $g,h \in \mathrm{GL}(2,\F_p)$ satisfy $gh = -hg$ and $\det(g) = \det(h) = 1$, as in part (a) of Proposition 14.4.4. Also assume that $p>2$.
\be
\item[(a)] Prove that the subgroup $\langle g,hangle \subset \mathrm{GL}(2,\F_p)$ is isomorphic to the quaternion group $Q = \{\pm 1, \pm i, \pm j, \pm k\}$, where $i^2 = j^2 = k^2 = -1, ij = -ji = k$, and $-1 \in Z(Q)$.
\item[(b)] Prove that $(M_3)_0$ is the normalizer of $\langle g,h angle$ in $\mathrm{GL}(2,\F_p)$.
\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}}

\begin{proof}
\item[(a)] The given conditions on $Q$ give the following Cayley's table of the quaternion group:
\begin{center}
$
\begin{array}{c|cccccccc|}
  	imes &1  &-1 &i &-i &j &-j &k &-k   \
  \hline
         1 &1  &-1 &i &-i &j &-j &k &-k    \
   	-1 &-1  &1 &-i &i &-j &j &-k &k    \
	i &i  &-i  &-1 &1 &k &-k &-j & j   \
	-i &-i  &i &1 &-1 &-k &k &j &-j    \
	j &j  &-j &-k &k &-1 &1 &i &-i    \
	-j &-j  &-j &k&-k &1 &-1&-i &i    \
	k &k  &-k &j &-j &-i &i &-1 &1    \
	-k &-k  &k &-j &j &i &-i &1 &-1    \
	\hline
\end{array}
$
\end{center}
Assume that $g,h \in \mathrm{GL}(2,\F_p)$ satisfy $gh = -hg$ and $\det(g) = \det(h) = 1$, as in part (a) of Proposition 14.4.4. ($p>2$). In the proof of this proposition, we have seen that
$$g^2 = -I_2,\qquad h^2 = -I_2.$$ 
Write $l = gh$. Then $l^2 = ghgh = -g^2h^2= -I_2$. Moreover $-I_2 \in \Z(\mathrm{GL}(2,\F_p))$. These properties give the table of $G = \{\pm I_2,\pm g, \pm h, \pm k\}$: 
\begin{center}
$
\begin{array}{c|cccccccc|}
  	imes &I_2  &-I_2 & g &-g &h &-h &l &-l   \
  \hline
         I_2 &I_2  &-I_2 &g &-g &h &-h &l &-l    \
   	-I_2 &-I_2  &I_2 &-g &g&-h &h &-l &l    \
	g&g  &-g  &-I_2 &I_2 &k &-k &-h & h   \
	-g &-g  &g &I_2 &-I_2 &-k &k &h&-h    \
	h &h  &-h &-l &l &-I_2 &I_2&g &-g    \
	-h &-h  &-h &l&-l &I_2 &-I_2&-g &g    \
	l &l  &-l &h &-h &-g &g &-I_2 &I_2    \
	-l &-l  &l &-h &h&g &-g &I_2 &-I_2    \
	\hline
\end{array}
$
\end{center}
Thus $G
e \varnothing$ is stable for multiplication, and each element has an inverse. Therefore $G$ is a subgroup of $\mathrm{GL}(2,\F_p)$. Moreover the two preceding tables show that the map $\phi : Q 	o G$ defined by $$\phi(1) = I_2, \phi(-1) = -I_2, \phi(i) = g, \phi(-i) = -g, \phi(j) = h, \phi(-j) = h, \phi(k) = l, \phi(-k) = -l,$$ is a group isomorphism.

Finally $\{g , h\}  \subset G$, and $G$ is a group, therefore $\langle g,h angle \subset G$. Conversely, since $-I_2 =g^2, l = gh$, every element of $G$ is generated by $g,h$. Therefore
$$\langle g,h angle = G = \{\pm I_2,\pm g, \pm h, \pm k\},$$
and 
$$\langle g,h angle \simeq Q.$$

Example with Sage:
\begin{verbatim}
F = GF(5)
A = matrix(F,2,[0,-1,1,0])
B = matrix(F,2,[2,0,0,-2])
gens = [A,B]
gens
\end{verbatim}
$$
\left[\left(\begin{array}{rr}
0 & 4 \
1 & 0
\end{array}ight), \left(\begin{array}{rr}
2 & 0 \
0 & 3
\end{array}ight)ight]
$$
\begin{verbatim}
G = MatrixGroup(gens)
G.list()
\end{verbatim}
$$
\left(\left(\begin{array}{rr}
0 & 1 \
4 & 0
\end{array}ight), \left(\begin{array}{rr}
0 & 2 \
2 & 0
\end{array}ight), \left(\begin{array}{rr}
0 & 4 \
1 & 0
\end{array}ight), \left(\begin{array}{rr}
0 & 3 \
3 & 0
\end{array}ight), \left(\begin{array}{rr}
1 & 0 \
0 & 1
\end{array}ight), \left(\begin{array}{rr}
2 & 0 \
0 & 3
\end{array}ight), \left(\begin{array}{rr}
4 & 0 \
0 & 4
\end{array}ight), \left(\begin{array}{rr}
3 & 0 \
0 & 2
\end{array}ight)ight)
$$
\begin{verbatim}
G.structure_description()
\end{verbatim}
\begin{center}
Q8
\end{center}

\item[(b)] Let 
$$\pi : 
\left \{
\begin{array}{ccc}
\mathrm{GL}(2,\F_p) & 	o & \mathrm{PGL}(2,\F_p)\
1 & \mapsto & [A].
\end{array}
ight.
$$
By definition, $(M_3)_0 = \pi^{-1}(N(H))$, where we identified $A \in \mathrm{GL}(2,\F_p)$ with $\gamma_{A,0} \in \mathrm{AGL}(2,\F_p)$.
Therefore, for all $A \in \mathrm{GL}(2,\F_p)$,
\begin{align*}
A \in (M_3)_0 &\iff [A] \in N(H) = N(\langle [g],[h] angle)\
&\iff 
\left\{
\begin{array}{lll}
{[}A{]} {[}g{]} {[}A{]}^{-1} &\in& \langle {[}g{]},{[}h{]}angle,\
{[}A{]}{[}h{]} {[}A{]}^{-1} &\in& \langle {[}g{]},{[}h{]} angle.
\end{array}
ight.
\end{align*}
Note that, if $[f] \in \langle {[}g{]},{[}h{]}angle = \{[I_2], [g], [h], [gh]\}$, where we can take $f \in \mathrm{SL}(2,\F_p)$ (since $I_2,g,h,gh \in  \mathrm{SL}(2,\F_p)$),
$$[A] [g] [A]^{-1} = [f] = \iff [AgA^{-1}] = [f] \iff AgA^{-1} = \lambda f,\ \lambda \in \F_p^*.$$
In this case, since $\det(g) = \det(h) = \det(f) = 1$,
$$1 = \det(AgA^{-1}) = \det(\lambda f) = \lambda^2 ,$$
thus $\lambda = \pm 1$, and $g = \pm f$. Therefore
\begin{align*}
[A][g][A]^{-1} \in \langle {[}g{]},{[}h{]}angle & \iff [A][g][A^{-1}] = [f], [f] \in \langle {[}g{]},{[}h{]}angle\
&\iff AgA^{-1} = \pm f, \ f \in \langle g,h angle\
&\iff AgA^{-1} \in  \langle g,h angle,
\end{align*}
We have the same result with $[A]h[A]^{-1}$, therefore, for all $A \in \mathrm{GL}(2,\F_p)$,
$$A \in (M_3)_0  \iff 
\left\{
\begin{array}{lll}
AgA^{-1} \in  \langle g,h angle\ 
AhA^{-1} \in  \langle g,h angle\ 
\end{array}
ight.
\iff A \in N_{\mathrm{GL}(2,\F_p)}(\langle g,h angle)
$$
This equivalence proves that 
$$(M_3)_0 = N_{\mathrm{GL}(2,\F_p)}(\langle g,h angle).$$
\end{proof}

Exercise 14.4.20

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Exercise 14.4.20

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