Let \(G\) be the group of \(2\times2\) determinant-one matrices over the four-element field \(\mathbb F_4\), and let \(S\) be the set of one-dimensional subspaces of \(\mathbb F_4^2\). Show that the natural action of \(G\) on \(S\) is faithful.
Solution
Let \[ e_1=(1,0)^T, \qquad e_2=(0,1)^T \] be the standard basis of \(\FF_4^2\).
1. If \(T\in G\) fixes every line in \(S\), then\[ T=\lambda I \]for some \(\lambda\in\FF_4^\times\).
Since \(T\) fixes the lines \(\FF_4e_1\) and \(\FF_4e_2\), there are \(a,b\in\FF_4^\times\) such that \[ Te_1=ae_1, \qquad Te_2=be_2. \] The line \[ \FF_4(e_1+e_2) \] is also fixed, so there is \(c\in\FF_4^\times\) with \[ T(e_1+e_2)=c(e_1+e_2). \] But linearity gives \[ T(e_1+e_2)=ae_1+be_2. \] Since \(e_1,e_2\) are linearly independent, \[ a=b=c. \] Thus \(T=aI\).
2. The only scalar matrix in \(G\) is the identity.
Let \(\lambda I\in G\). Since its determinant is \(1\), \[ \lambda^2=1. \] The field \(\FF_4\) has characteristic \(2\), so \[ (\lambda-1)^2 = \lambda^2-2\lambda+1 = 0. \] A field has no nonzero nilpotents, hence \[ \lambda-1=0. \] Therefore \(\lambda=1\) and \(\lambda I=I\).
3. The kernel of the action of \(G\) on \(S\) is trivial.
An element of the kernel fixes every line in \(S\). By step 1 it is scalar, and by step 2 it is the identity. Hence the kernel consists only of \(I\).
4. The natural action of \(G\) on \(S\) is faithful.
An action is faithful exactly when its kernel is trivial. This is step 3.
5. Q.E.D.
Step 4 is the required conclusion.