Let’s look at \(\SL_2(\FF_3)\). How many elements are in that group? What is its centre? Identify \(\PSL_2(\FF_3)\) as a permutation group.
Solution
1. The group \(\operatorname{SL}_2(\FF_3)\) has order \(24\).
First \[ |\operatorname{GL}_2(\FF_3)|=(3^2-1)(3^2-3)=8\cdot6=48. \] The determinant map \[ \det:\operatorname{GL}_2(\FF_3)\to\FF_3^\times \] is surjective, and \(|\FF_3^\times|=2\). Its kernel is \(\operatorname{SL}_2(\FF_3)\), so \[ |\operatorname{SL}_2(\FF_3)|=48/2=24. \]
2. The center of \(\operatorname{SL}_2(\FF_3)\) is\[ Z(\operatorname{SL}_2(\FF_3))=\{I,-I\}. \]
Let \[ A=\begin{pmatrix}a&b\\c&d\end{pmatrix} \] be central. It commutes with the two elementary matrices \[ U=\begin{pmatrix}1&1\\0&1\end{pmatrix}, \qquad L=\begin{pmatrix}1&0\\1&1\end{pmatrix} \] in \(\operatorname{SL}_2(\FF_3)\). From \(AU=UA\) one gets \(c=0\) and \(a=d\); from \(AL=LA\) one then gets \(b=0\). Thus \(A=aI\) is scalar. The determinant condition gives \[ a^2=1 \] in \(\FF_3\), so \(a=\pm1\). Hence the center is exactly \(\{\pm I\}\).
3. Therefore\[ |\operatorname{PSL}_2(\FF_3)|=12. \]
By definition, \[ \operatorname{PSL}_2(\FF_3)=\operatorname{SL}_2(\FF_3)/\{\pm I\}. \] Use 1 and 2.
4. The natural action on the projective line \(\PP^1(\FF_3)\) gives an embedding\[ \operatorname{PSL}_2(\FF_3)\hookrightarrow S_4. \]
2. Thus the induced action of \(\operatorname{PSL}_2(\FF_3)\) is faithful.
5. Under this action,\[ \operatorname{PSL}_2(\FF_3)\cong A_4. \]
By <1>3–<1>4, the image is a subgroup of \(S_4\) of order \(12\), hence of index \(2\). Any index-\(2\) subgroup is the kernel of a surjective homomorphism \(S_4\to C_2\); the unique such subgroup is \(A_4\). Therefore the image is \(A_4\).