Exercise 14.3.15

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}}

Prove that $\mathrm{GL}(2,\F_2) = \mathrm{SL}(2,\F_2) \simeq \mathrm{PSL}(2,\F_2) \simeq S_3$, and ${\mathrm{PSL}(2,\F_3) \simeq A_4}$.

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}
 Since $\mathrm{SL}(2,\F_2)$ is the kernel of the surjective homomorphism $det : \mathrm{GL}(2,\F_2) 	o \F_2^*$, where $\F_2^* = \{1\}$ is trivial, we obtain $\mathrm{SL}(2,\F_2) = \mathrm{GL}(2,\F_2)$. 
 
 By part (c) of Exercise 14, $\mathrm{PGL}(2,\F_q) \simeq \mathrm{SL}(2,\F_q)/\{I_2,-I_2\}.$
   For $q = 2$, $I_2 = -I_2$, thus 
  $$\mathrm{PGL}(2,\F_2) \simeq \mathrm{SL}(2,\F_2) = \mathrm{GL}(2,\F_2).$$
  Moreover,
  $$\mathrm{GL}(2,\F_2) = \left\{
   \begin{pmatrix}
  1 & 0\
  0 & 1
  \end{pmatrix},
  \begin{pmatrix}
  0 & 1\
  1 & 1
  \end{pmatrix},
   \begin{pmatrix}
  0 & 1\
  1 & 0
  \end{pmatrix},
   \begin{pmatrix}
  1 & 0\
  1 & 1
  \end{pmatrix},
   \begin{pmatrix}
  1 & 1\
  0 & 1
  \end{pmatrix},
    \begin{pmatrix}
  1 & 1\
  1 & 0
  \end{pmatrix}
  ight\},
  $$
  thus $|\mathrm{GL}(2,\F_2)| = 6$, and $\mathrm{GL}(2,\F_2) $ is not Abelian:
  $$
   \begin{pmatrix}
  1 & 1\
  1 & 0
  \end{pmatrix}
  \begin{pmatrix}
  1 & 1\
  0 & 1
  \end{pmatrix}
  =
    \begin{pmatrix}
  1 & 0\
  1 & 1
  \end{pmatrix}
  
e
    \begin{pmatrix}
  0 & 1\
  1 & 0
  \end{pmatrix}
=
 \begin{pmatrix}
  1 & 1\
  0 & 1
  \end{pmatrix}
  \begin{pmatrix}
  1 & 1\
  1 & 0
  \end{pmatrix}.
  $$
Since there is only one non Abelian group of order 6, up to isomorphism,
  $$\mathrm{GL}(2,\F_2) = \mathrm{SL}(2,\F_2) \simeq \mathrm{PSL}(2,\F_2) \simeq S_3.$$
  
  \bigskip
  
To prove the last inclusion, we use the action of $\mathrm{PGL}(2,F)$ on $F \cup \{\infty\}$ described in section 7.5. We resume this description in the following lemma:

{\bf Lemma. }{\it The operation defined for all  $M = 
\left(
\begin{array}{cc}
 a&   b  \
 c&   d   
\end{array}
ight)$ and $z \in \dot F$ by 
$$
[M] \cdot z = \frac{az+b}{cz+d}, \qquad (z \in F \setminus \{-d/c\}),
$$
and also
$$ [M] \cdot (-d/c) = \infty, \qquad [M] \cdot \infty = a/c \qquad ([M] \cdot \infty = \infty 	ext { if } c = 0).$$
is a (left) action of $\mathrm{PGL}(2, F)$ on the projective line $\dot F$.
}

\bigskip

In this exercise,  $F = \F_3$. Consider the restriction of this action to the subgroup $\mathrm{PSL}(2,\F_3)$ on $\dot{\F}_3= \F_3 \cup \{\infty\} = \{0,1,2,\infty\}$.

This action defines a group homomorphism
 $$\varphi
 \left\{
 \begin{array}{ccc}
 \mathrm{PSL}(2,\F_3) & 	o & S(\dot{\F}_3)\
 {[}M{]} & \mapsto & 
     \varphi([M]) \left \{
     \begin{array}{ccc}
      \dot{\F}_3 & 	o & \dot{\F}_3\
      z & \mapsto &[M]\cdot z.
     \end{array}
     ight.
  \end{array}
 ight.
 $$
 We show that $\ker(\varphi)$ is trivial (the action is faithful): Let $[M] \in  \mathrm{PSL}(2,\F_3)$, with $M  =\left(
\begin{array}{cc}
 a&   b  \
 c&   d   
\end{array}
ight) \in  \mathrm{SL}(2,\F_3)$, such that $\varphi([M])$ is the identity of $\dot \F_3$. Then, for all $z \in \dot \F_3, [M]\cdot z = z$. The equality $[M]\cdot \infty = \infty$ shows that $c=0$. Thus for every $z\in \F_3$,
$$\frac{az+b}{d} = z,$$
and then $az + b = dz, (a-d)z + b = 0$. With $z = 0$, we obtain $b=0$, and with $z=1$, $a-d = 0$. Therefore $M  = a I_2, \ a \in \F_3$, and $[M] = [I_2]$. We have proved that the kernel of the action is trivial, in other words $\varphi$ is injective, and $ \mathrm{PSL}(2,\F_3) \simeq \im(\varphi) \subset S(\dot \F_3)$ is isomorphic to a subgroup of  $S(\dot \F_3) \simeq S_4$, therefore $  \mathrm{PSL}(2,\F_3)$ is isomorphic to a subgroup of $S_4$. By Exercise 14, parts (c) and (d), $| \mathrm{PSL}(2,\F_3)| = 12$, and the only subgroup of $S_4$ of order $12$ is $A_4$. Therefore
$$ \mathrm{PSL}(2,\F_3) \simeq A_4.$$

\end{proof}
  
 Note:  We explain why the permutations corresponding to elements of $ \mathrm{PSL}(2,\F_3) \simeq A_4$ are even, with another proof of this isomorphism.
 
 Let $[T],[S]$ the two elements of $\mathrm{PSL}(2,\F_3)$ defined by $[T ]\cdot z = z+1$ and $[S]\cdot z = -1/z$, corresponding to the matrices 
 $$T = 
 \begin{pmatrix}
  1 & 1\
   0 & 1
  \end{pmatrix},
  \qquad 
  S = 
   \begin{pmatrix}
  0 & -1\
   1 & 0
  \end{pmatrix},
  \qquad T,S \in \mathrm{SL}(2,\F_3).
$$
The permutation of $\dot \F_3$ corresponding to $[T],[S]$ by $\varphi$ are
$$\varphi([T]) = 
\left(
\begin{array}{cccc}
0 & 1 & 2 & \infty\
1 & 2 & 0 & \infty
\end{array}
ight) =(0\,1\,2), \qquad 
\varphi([S]) = \left(
\begin{array}{cccc}
0 & 1 & 2 & \infty\
\infty & 2 & 1 & 0
\end{array}
ight) = (0\,\infty)(1\, 2)
$$
If we take a numbering of $\dot \F_3$ to $\{1,2,3,4\}$, for instance $ f : 0 \mapsto 1, 1 \mapsto 2, 2 \mapsto 3, \infty \mapsto 4$, the permutations corresponding to $T,S$ are 
$(1\,2\,3) , (1\,4)(2\,3)$, both even.

A well known theorem shows that the matrices $[T],[S]$ generate $\mathrm{PSL}(2,\Z)$ : every matrix $M \in \mathrm{SL}(2,\Z)$ is of the form
$$M = \pm U_1\cdots U_k,\quad U_i \in \{T,S\}.$$

Reducing modulo 3, every matrix $M \in \mathrm{SL}(2,\F_3)$ is of the same form, thus $[T],[S]$ generate $\mathrm{PSL}(3,\F_3).$ This proves that the permutations corresponding to  $\mathrm{PSL}(3,\F_3)  = \langle [S], [T] angle$ are all even, and the argument of cardinality shows that $\mathrm{PSL}(3,\F_3) \simeq \im(\varphi) \simeq A_4$.

With Sage:
 \begin{verbatim}
 G = PermutationGroup(['(1,4)(2,3)','(1,2,3)'])
 G.order()
 		12
 G.structure_description()
	        'A4'		
 \end{verbatim}
 
 This confirms $A_4 = \langle (1\,4)(2\, 3), (1\,2\,3) angle$, and consequently $\mathrm{PSL}(3,\F_3)  = \langle [S], [T] angle$.
 
 \bigskip
 
 Note 2: The same method can be applied to prove anew the first isomorphism $\mathrm{PSL}(2,\F_2) \simeq S_3$. Indeed, the action of $\mathrm{PGL}(2,\F_2)$ on $\F_2 \cup \{\infty\} = \{0,1,\infty\}$, restricted to $\mathrm{PSL}(2,\F_2)$, gives an injective group homomorphism
 
 $$\varphi
 \left\{
 \begin{array}{ccc}
 \mathrm{PSL}(2,\F_2) & 	o & S(\dot{\F}_2)\simeq S_3\
 {[}M{]} & \mapsto & 
     \varphi([M]) \left \{
     \begin{array}{ccc}
      \dot{\F}_2 & 	o & \dot{\F}_2\
      z & \mapsto &[M]\cdot z.
     \end{array}
     ight.
  \end{array}
 ight.
 $$
 
 Moreover $| \mathrm{PSL}(2,\F_2)| = |S(\dot{\F}_2)| = 6$, thus $\varphi$ is bijective. Therefore $\mathrm{PSL}(2,\F_2) \simeq S_3$.

Exercise 14.3.15

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Exercise 14.3.15

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