For \(n = 3, 4\), and 5, determine the number of similarity classes of matrices \(A \in M_n(\mathbb{Q})\) such that \(A^8 = I\) but \(A^4 \neq I\).
Solution
The numbers of similarity classes for \(n=3,4,5\) are respectively \(0,1,2\).
1. The condition \(A^8=I\) gives a unique multiplicitydescription\[ \mathbb Q^n\cong E_1^{\oplus a}\oplus E_2^{\oplus b} \oplus E_4^{\oplus c}\oplus E_8^{\oplus d}, \qquad n=a+b+2c+4d, \]as a \(\mathbb Q[x]\)-module with \(x\) acting by \(A\), where\[ E_1=\mathbb Q[x]/(x-1),\quad E_2=\mathbb Q[x]/(x+1), \quad E_4=\mathbb Q[x]/(x^2+1),\quad E_8=\mathbb Q[x]/(x^4+1). \]
Factor \[ x^8-1=(x-1)(x+1)(x^2+1)(x^4+1). \] The two linear factors are irreducible. The quadratic has no rational root. Substituting \(x=t+1\) in the quartic gives \(t^4+4t^3+6t^2+4t+2\), which is Eisenstein at \(2\), so the quartic is irreducible as well [@DF04]. These are distinct monic irreducibles, so the displayed factorization is squarefree.
The module is finitely generated, since a finite rational basis also generates over \(\mathbb Q[x]\), and is annihilated by \(x^8-1\). The structure theorem over the PID \(\mathbb Q[x]\) expresses it uniquely as a direct sum of modules \(\mathbb Q[x]/(p^e)\) for monic irreducibles \(p\) [@DF04]. On such a summand, annihilation by \(x^8-1\) implies \(p^e\mid x^8-1\) by applying that polynomial to the class of \(1\). Hence \(e=1\) and \(p\) is one of the four displayed factors. Their degrees give the dimension equation. Conversely, every direct sum of this form is annihilated by \(x^8-1\).
An invertible rational linear map intertwines the two matrix actions exactly when it is a \(\mathbb Q[x]\)-module isomorphism. Thus uniqueness of the multiplicities is exactly uniqueness of the similarity class.
2. The condition \(A^4\ne I\) is equivalent to \(d\geq1\).
On \(E_1,E_2,E_4\), the polynomial \(x^4-1\) acts by zero, since their defining polynomials divide it. On \(E_8\) one has \(x^4=-1\), so \(x^4-1\) acts as multiplication by \(-2\). This is nonzero on every nonzero copy of \(E_8\) over \(\mathbb Q\). Therefore \(A^4-I\) is nonzero exactly when at least one such summand occurs.
3. The dimensions \(3,4,5\) yield exactly the stated counts.
For \(n=3\), the inequalities \(d\geq1\) and \(4d\leq n\) are incompatible, so there are no such matrices. For \(n=4\), the only possibility is \((a,b,c,d)=(0,0,0,1)\). For \(n=5\), one must have \(d=1\) and \(a+b+2c=1\). Thus \(c=0\) and \((a,b)=(1,0)\) or \((0,1)\). These cases exhaust the nonnegative integer solutions.
For explicit representatives, put \[ C=\begin{pmatrix} 0&0&0&-1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{pmatrix}. \] This is multiplication by \(x\) on \(E_8\) in the basis \(1,x,x^2,x^3\), so \(C^4=-I_4\) and \(C^8=I_4\). It represents the single class in dimension four. In dimension five, take \(\operatorname{diag}(C,1)\) and \(\operatorname{diag}(C,-1)\). Both have eighth power the identity and fourth power different from the identity. They are not similar, since their characteristic polynomials are respectively \((x^4+1)(x-1)\) and \((x^4+1)(x+1)\). Step 1 proves that there are no additional classes.