Exercise 4.5.11 ($\mathrm{SL}_2(\mathbb{F}_3)/ Z(\mathrm{SL}_2(\mathbb{F}_3) ) \simeq A_4$)

Proof. Let $G$ be the group ${\mathrm{SL}}_2({𝔽}_3)$, of order $24$.

By Exercise 2.4.9,

where

Let $Z=Z(G)$ be the center of $G$. Of course, $I\in Z$ and $-I\in Z$. Conversely, let $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ be any element of $Z$, where $a,b,c,d\in {𝔽}_3$ satisfy $\det <!--\; FUNCTION\; APPLICATION\; -->(M)=\mathit{𝑎𝑑}-\mathit{𝑏𝑐}=1$. Then

Thus $M$ is of the form $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)$. Then

Therefore $M=\mathit{𝑎𝐼}$, where $\det <!--\; FUNCTION\; APPLICATION\; -->(M)=a^2=1$, thus $a=\pm 1$. This shows that

Since $Z⊴G$, the quotient group ${\mathrm{PSL}}_2({𝔽}_3)={\mathrm{SL}}_2({𝔽}_3)∕Z({\mathrm{SL}}_2({𝔽}_3))$ is well defined, and

Now we prove $\mathrm{\Gamma }={\mathrm{SL}}_2({𝔽}_3)∕Z({\mathrm{SL}}_2({𝔽}_3))\simeq A_4$.

First solution (following the hint.)

By the Example (Groups of order 12 p. 144) we know that $\mathrm{\Gamma }$ has a normal Sylow $3$-subgroup, or $\mathrm{\Gamma }\simeq A_4$.

Assume for the sake of contradiction that $\mathrm{\Gamma }$ has a normal Sylow $3$-subgroup $\bar{H}$, of order $3$. Consider the natural projection $\pi :G\to G∕Z=\mathrm{\Gamma }$. Then $H={\pi }^{-1}(\bar{H})$ is a normal subgroup of $G={\mathrm{SL}}_2({𝔽}_3)$ of order $6$. Since $|H|=6$, $H$ has a unique subgroup $K$ of order $3$ ( $n_3(H)=1$). So $K$ is a characteristic subgroup of $H$, where $H⊴G$, therefore $K⊴G$. So ${\mathrm{SL}}_2({𝔽}_3)$ has a unique Sylow $3$-subgroup $K$. This is in contradiction with Exercise 9, which shows that ${\mathrm{SL}}_2({𝔽}_3)$ has four subgroups of order $3$.

This contradiction shows that $\mathrm{\Gamma }$ has no normal Sylow $3$-subgroup. Therefore $\mathrm{\Gamma }\simeq A_4$:

Second solution (without Sylow’s Theorem). The vector plane $P={𝔽}_3\times {𝔽}_3$ has exactly four lines, generated by $(1,0),(1,1),(1,2)$ and $(0,1)$ (corresponding to the four points of the projective line ${\mathbb{P}}_1({𝔽}_3)$). Consider the set $S$ of these $4$ lines. The group ${\mathrm{GL}}_2({𝔽}_3)$ acts on the plane $P$ by the action defined by $A\cdot (a,b)=A\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right),$ and also its subgroup $G={\mathrm{SL}}_2({𝔽}_3)$. Since $A\cdot (\mathit{𝜆𝑎},\mathit{𝜆𝑏})=\mathit{𝜆𝐴}\cdot (a,b)$, this action induces an action of $G$ on the set $S$ of the four lines of $P$, so gives a permutation representation

Let $K$ be the kernel of this representation. Then $I\in K$ and $-I\in K$ since these two matrices preserve each line. Let $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$be any element of $K$. Since $M$ preserves each line, there are some scalars ${\lambda }_1,{\lambda }_2,{\lambda }_3\in {𝔽}_3$ such that

which give

thus $b=c=0$, and $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right)$. Finally,

thus $a=d$, so $M=\mathit{𝑎𝐼}$, and $\det <!--\; FUNCTION\; APPLICATION\; -->(M)=a^2=1$, thus $a=\pm 1$. We obtain

By the First Isomorphism Theorem, $\mathrm{\Gamma }={\mathrm{SL}}_2({𝔽}_3)∕Z({\mathrm{SL}}_2({𝔽}_3))$ is isomorphic to a subgroup $L$ of $S_4$. Moreover $|L|=|\mathrm{\Gamma }|=12=|A_4|$, and $A_4$ is the unique subgroup of $S_4$ of order $12$, therefore $L=A_4$, and

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