Exercise 4.5.41 ($\mathrm{SL}_2(\mathbb{F}_4)\simeq A_5$)

Proof. There is field with $4$ elements, unique up to isomorphism. Since $x^2+x+1$ is irreducible over ${𝔽}_2$, ${𝔽}_4$ can be defined by ${𝔽}_4\simeq {𝔽}_2[X]∕(x^2+x+1){𝔽}_2[X]$. If $\alpha$ is the class of $x$ modulo $x^2+x+1$, then

The vector plane $P={𝔽}_4\times {𝔽}_4$ has exactly five lines, generated by $(1,\gamma )$ for $\gamma \in {𝔽}_4$ and also $(0,1)$. Consider the set $S$ of these $5$ lines. The group ${\mathrm{GL}}_2({𝔽}_4)$ 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 by restriction its subgroup $G={\mathrm{SL}}_2({𝔽}_4)$ acts on $P$. 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 affords a permutation representation

Let $K$ be the kernel of this representation. Then $I\in K$ (note that $-I=I$ since the characteristic of ${𝔽}_4$ is $2$). 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 $\det <!--\; FUNCTION\; APPLICATION\; -->(M)=a^2=1$. Since ${\alpha }^2=1+\alpha \ne 1$ and ${({\alpha }^2)}^2=\alpha \ne 1$, then $a=1$. We obtain

so the action of ${\mathrm{SL}}_2({𝔽}_4)$ on the set $S$ is faithful, and ${\mathrm{SL}}_2({𝔽}_4)$ is isomorphic to a subgroup of $S_5$:

For $q=4$,

Consider now the map

where ${𝔽}_4^{\times }$ is a group of order $3$.

Then

• $\phi$ is a homomorphism, because $\phi (\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(A)\det <!--\; FUNCTION\; APPLICATION\; -->(B)=\phi (A)\phi (B)$.
• $\phi$ is surjective : if $\lambda \in {𝔽}_4^{\ast }$, then $\det <!--\; FUNCTION\; APPLICATION\; -->(M)=\lambda$, where $M=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in {\mathrm{GL}}_2({𝔽}_4)$.

• Let $A\in {\mathrm{GL}}_2({𝔽}_4)$. Then $A\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ if and only if $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=1$, so

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )={\mathrm{SL}}_2({𝔽}_4).$$

The first isomorphism Theorem gives ${\mathrm{GL}}_2({𝔽}_4)∕{\mathrm{SL}}_2({𝔽}_4)\simeq {𝔽}_4^{\times }$, thus

So $|{\mathrm{SL}}_2({𝔽}_4)|=|A_5|$, and by (1), ${\mathrm{SL}}_2({𝔽}_4)$ is isomorphic to a subgroup of $A_5$, therefore

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