Suppose \(A \in M_4(\mathbb{Q})\) is a 4-by-4 matrix of multiplicative order 6. We use \(A\) to give \(\mathbb{Q}^4\) the structure of a \(\mathbb{Q}[x]\)-module in the usual way, by setting \(x \cdot v = Av\) for \(v \in \mathbb{Q}^4\).
What are all of the isomorphism classes of \(\mathbb{Q}[x]\)-modules that can arise this way?
For each of your answers above, write down the rational canonical form of the corresponding matrix \(A\).
What are all the of the abelian groups of size 24 (up to isomorphism)?
Solution
Over \(\QQ\), \[ x^6-1=\Phi_1\Phi_2\Phi_3\Phi_6=(x-1)(x+1)(x^2+x+1)(x^2-x+1), \] a product of distinct monic irreducibles. For a monic polynomial \(p\), write \(C(p)\) for its companion matrix, with ones on the subdiagonal and last column the negated lower coefficients of \(p\).
1. The \(\QQ[x]\)-module \(V=\QQ^4\) is isomorphic to\[ V_{a}=\bigoplus_{d\in\{1,2,3,6\}}\bigl(\QQ[x]/(\Phi_d)\bigr)^{a_d}, \qquad a_1+a_2+2a_3+2a_6=4, \]for unique multiplicities \(a_d\ge0\), and \(A\) has order exactly \(6\) if and only if \(a_6\ge1\), or \(a_2\ge1\) and \(a_3\ge1\).
Since \(A^6=I\), the module is annihilated by the squarefree polynomial \(x^6-1\), so by the structure theorem over the PID \(\QQ[x]\) it is a direct sum of modules \(\QQ[x]/(\Phi_d)\) with \(d\mid6\), with unique multiplicities [@DF04]; dimensions give \(a_1+a_2+2a_3+2a_6=4\). On \(\QQ[x]/(\Phi_d)\), \(x\) acts with order exactly \(d\), so the order of \(A\) is the least common multiple of the \(d\) with \(a_d\ge1\). This least common multiple is \(6\) exactly when \(6\) occurs, or both \(2\) and \(3\) occur.
2. For part (a), there are exactly seven classes, listed by \((a_1,a_2,a_3,a_6)\):\[ (0,0,0,2),\ (2,0,0,1),\ (0,2,0,1),\ (1,1,0,1),\ (0,0,1,1),\ (1,1,1,0),\ (0,2,1,0). \]
If \(a_6=2\), the others vanish. If \(a_6=1\), then \(a_1+a_2+2a_3=2\), with the four solutions \((2,0,0)\), \((1,1,0)\), \((0,2,0)\), \((0,0,1)\). If \(a_6=0\), then \(a_2,a_3\ge1\) and \(a_1+a_2+2a_3=4\) force \(a_3=1\) and \((a_1,a_2)\in\{(1,1),(0,2)\}\). By step 1 these seven modules are pairwise nonisomorphic and are exactly the modules that arise.
3. For part (b), the rational canonical forms are given by the invariant factors in the table.
| \((a_1,a_2,a_3,a_6)\) | Invariant factors | Rational canonical form |
|---|---|---|
| \((0,0,0,2)\) | \(x^2-x+1,\ x^2-x+1\) | \(\operatorname{diag}\bigl(C(x^2-x+1),C(x^2-x+1)\bigr)\) |
| \((2,0,0,1)\) | \(x-1,\ x^3-2x^2+2x-1\) | \(\operatorname{diag}\bigl(1,C(x^3-2x^2+2x-1)\bigr)\) |
| \((0,2,0,1)\) | \(x+1,\ x^3+1\) | \(\operatorname{diag}\bigl(-1,C(x^3+1)\bigr)\) |
| \((1,1,0,1)\) | \(x^4-x^3+x-1\) | \(C(x^4-x^3+x-1)\) |
| \((0,0,1,1)\) | \(x^4+x^2+1\) | \(C(x^4+x^2+1)\) |
| \((1,1,1,0)\) | \(x^4+x^3-x-1\) | \(C(x^4+x^3-x-1)\) |
| \((0,2,1,0)\) | \(x+1,\ x^3+2x^2+2x+1\) | \(\operatorname{diag}\bigl(-1,C(x^3+2x^2+2x+1)\bigr)\) |
Explicitly, \[ C(x^2-x+1)=\begin{pmatrix}0&-1\\1&1\end{pmatrix},\quad C(x^3-2x^2+2x-1)=\begin{pmatrix}0&0&1\\1&0&-2\\0&1&2\end{pmatrix},\quad C(x^3+1)=\begin{pmatrix}0&0&-1\\1&0&0\\0&1&0\end{pmatrix}, \] \[ C(x^3+2x^2+2x+1)=\begin{pmatrix}0&0&-1\\1&0&-2\\0&1&-2\end{pmatrix},\quad C(x^4-x^3+x-1)=\begin{pmatrix}0&0&0&1\\1&0&0&-1\\0&1&0&0\\0&0&1&1\end{pmatrix}, \] \[ C(x^4+x^2+1)=\begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&-1\\0&0&1&0\end{pmatrix},\quad C(x^4+x^3-x-1)=\begin{pmatrix}0&0&0&1\\1&0&0&1\\0&1&0&0\\0&0&1&-1\end{pmatrix}. \]
The invariant factors are obtained by multiplying, from the largest down, one factor \(\Phi_d\) for each \(d\) with \(a_d\) remaining; for instance \((0,2,1,0)\) gives \(\Phi_2\) and \(\Phi_2\Phi_3=(x+1)(x^2+x+1)=x^3+2x^2+2x+1\). The rational canonical form is the block diagonal matrix of the companion matrices of the invariant factors [@DF04].
4. For part (c), the abelian groups of order \(24\) are \(\ZZ/24\), \(\ZZ/12\times\ZZ/2\), and \(\ZZ/6\times(\ZZ/2)^2\).
Since \(24=2^3\cdot3\), an abelian group of order \(24\) is the product of its Sylow \(2\)-subgroup of order \(8\) and its Sylow \(3\)-subgroup \(\ZZ/3\) [@DF04]. The partitions \(3\), \(2+1\), \(1+1+1\) give the Sylow \(2\)-subgroups \(\ZZ/8\), \(\ZZ/4\times\ZZ/2\), \((\ZZ/2)^3\). By the Chinese remainder theorem the products with \(\ZZ/3\) are the three listed groups.
5. Q.E.D.
Steps 2, 3, and 4 answer parts (a), (b), and (c).