Let \(p\) be prime. Show that every element of \[ GL_2(\mathbb F_p) \] has order dividing either \[ p^2-1 \qquad\text{or}\qquad p(p-1). \]
Solution
Let \[ A\in\GL_2(\FF_p). \]
1. If the characteristic polynomial of \(A\) is irreducible over\(\FF_p\), then the order of \(A\) divides\[ p^2-1. \]
In this case the characteristic polynomial is also the minimal polynomial, because both have degree \(2\). Therefore \[ \FF_p[A] \cong \FF_p[t]/(m_A(t)). \] Since \(m_A\) is irreducible quadratic, this quotient is the field with \(p^2\) elements: \[ \FF_p[A]\cong\FF_{p^2}. \] The matrix \(A\) is invertible, so it corresponds to an element of \[ \FF_{p^2}^{\times}, \] a group of order \(p^2-1\). By Lagrange’s theorem, the order of \(A\) divides \(p^2-1\).
2. Suppose the characteristic polynomial of \(A\) splits over \(\FF_p\)with two distinct roots. Then the order of \(A\) divides\[ p-1. \]
With distinct eigenvalues, \(A\) is diagonalizable over \(\FF_p\): \[ A\sim \begin{pmatrix} \lambda&0\\ 0&\mu \end{pmatrix}, \] where \[ \lambda,\mu\in\FF_p^\times. \] Each nonzero field element satisfies \[ \lambda^{p-1}=\mu^{p-1}=1. \] Hence \[ A^{p-1}=I, \] so the order divides \(p-1\).
3. Suppose the characteristic polynomial is\[ (t-\lambda)^2 \]with \(\lambda\in\FF_p^\times\). Then the order of \(A\) divides\[ p(p-1). \]
If \(A=\lambda I\), its order divides \(p-1\), so the conclusion follows. Otherwise its Jordan form over \(\FF_p\) is \[ A\sim \lambda(I+N), \qquad N= \begin{pmatrix} 0&c\\ 0&0 \end{pmatrix} \] for some \(c\neq0\), with \[ N^2=0. \] The scalar matrix \(\lambda I\) commutes with \(I+N\). Moreover, \[ \lambda^{p-1}=1 \] and, in characteristic \(p\), \[ (I+N)^p = I+pN = I, \] because every higher power of \(N\) vanishes. Therefore \[ A^{p(p-1)} = \lambda^{p(p-1)}(I+N)^{p(p-1)} = I. \] Thus the order of \(A\) divides \(p(p-1)\).
4. Every element of \(\GL_2(\FF_p)\) has order dividing either\[ p^2-1 \]or\[ p(p-1). \]
3. Since \(p-1\) divides \(p(p-1)\), thedistinct-root case also satisfies the second bound.
5. Q.E.D.
Step 4 is the required dichotomy.