Let \(G\) be the subgroup of \(\text{GL}_3(\mathbb C)\) generated by the three matrices \[\begin{align*} A= \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{pmatrix}, \quad B= \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}, \quad C= \begin{pmatrix} i & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \end{align*}\]
where \(i^2=-1\). Here \(\mathbb C\) denotes the complex field.
Compute the order of \(G\).
Find a matrix in \(G\) of largest possible order (as an element of \(G\)) and compute this order.
Compute the number of elements in \(G\) with this largest order.
Solution
1. The matrices \(A\) and \(B\) generate the permutation-matrix copy of \(S_3\) in \(\operatorname{GL}_3(\mathbb C)\).
The matrix \(A\) interchanges the first and third standard basis vectors and fixes the second, so it represents the transposition \((1\,3)\). The matrix \(B\) sends \[ e_1\mapsto e_2,\qquad e_2\mapsto e_3,\qquad e_3\mapsto e_1, \] so it represents the \(3\)-cycle \((1\,2\,3)\). A transposition and a \(3\)-cycle generate \(S_3\).
2. Conjugates of \(C\) by the permutation subgroup generate\[ D=\{\operatorname{diag}(i^a,i^b,i^c):a,b,c\in\mathbb Z/4\mathbb Z\} \cong C_4^3. \]
Conjugating \(C=\operatorname{diag}(i,1,1)\) by permutation matrices moves the entry \(i\) to any of the three diagonal positions. Thus \(G\) contains \[ \operatorname{diag}(i,1,1),\quad \operatorname{diag}(1,i,1),\quad \operatorname{diag}(1,1,i). \] These commuting matrices independently generate all diagonal matrices whose diagonal entries lie in \[ \mu_4=\{1,i,-1,-i\}. \] Hence they generate \(D\cong C_4^3\), which has order \(4^3=64\).
3. Every element of \(G\) has a unique expression\[ dP_\sigma, \qquad d\in D,\ \sigma\in S_3, \]and therefore\[ |G|=64\cdot6=384. \]
The subgroup \(D\) is normalized by permutation matrices, since conjugating a diagonal matrix merely permutes its diagonal entries. Thus the subgroup generated by \(D\) and the permutation matrices is the semidirect product \[ D\rtimes S_3. \] Their intersection is trivial: the only diagonal permutation matrix is the identity. Hence the expression \(dP_\sigma\) is unique, and \[ |G|=|D||S_3|=64\cdot6=384. \]
4. Let \(g=dP_\sigma\) and suppose \(\sigma\) has a cycle\[ (j_1\,j_2\,\dots\,j_\ell). \]If the corresponding diagonal entries of \(d\) are \(z_{j_1},\dots,z_{j_\ell}\in\mu_4\), then the restriction of \(g\) to those coordinates has order\[ \ell\,\operatorname{ord}(z_{j_1}\cdots z_{j_\ell}). \]
After \(\ell\) applications of \(g\), the permutation part returns each coordinate in the cycle to itself, and the accumulated scalar on every coordinate in that cycle is \[ s=z_{j_1}\cdots z_{j_\ell}. \] Thus the \(\ell\)th power of the cycle block is scalar multiplication by \(s\). If \(g^k\) is the identity on the block, then the permutation part forces \(\ell\mid k\), say \(k=\ell m\), and then \(s^m=1\). Hence the least such \(k\) is \[ \ell\,\operatorname{ord}(s). \]
5. Elements whose permutation part is the identity have order at most \(4\), and those whose permutation part is a transposition have order at most \(8\).
For the identity permutation, the element lies in \(D\), whose exponent is \(4\).
For a transposition, there is one cycle of length \(2\) and one fixed point. By 4, the \(2\)-cycle block has order at most \[ 2\cdot4=8, \] while the fixed coordinate contributes order at most \(4\). The order of the whole element is the least common multiple of these block orders, hence at most \(8\).
6. If the permutation part is a \(3\)-cycle, the element has order at most \(12\), and order \(12\) occurs exactly when the product of its three diagonal entries has order \(4\) in \(\mu_4\).
A \(3\)-cycle has a single cycle block of length \(3\). By 4, its order is \[ 3\,\operatorname{ord}(z_1z_2z_3). \] The product lies in \(\mu_4\), so its order is at most \(4\). Thus the element order is at most \(12\), with equality exactly when \[ z_1z_2z_3\in\{i,-i\}. \]
7. The matrix\[ CB= \begin{pmatrix} 0&0&i\\ 1&0&0\\ 0&1&0 \end{pmatrix} \]has order \(12\).
The action of \(CB\) on the standard basis is \[ e_1\mapsto e_2,\qquad e_2\mapsto e_3,\qquad e_3\mapsto i e_1. \] Hence \[ (CB)^3=iI_3. \] Therefore \((CB)^{12}=I_3\). Conversely, if \((CB)^k=I_3\), the permutation part forces \(3\mid k\), say \(k=3m\). Then \[ I_3=(CB)^{3m}=i^mI_3, \] so \(4\mid m\). Hence \(12\mid k\), and the order is exactly \(12\).
8. The largest possible element order in \(G\) is therefore\[ \boxed{12}. \]
By 5 and 6, every element has order at most \(12\). By 7, an element of order \(12\) exists.
9. Exactly \(64\) elements of \(G\) have order \(12\).
By 5, an element of order \(12\) must have permutation part a \(3\)-cycle. There are exactly two \(3\)-cycles in \(S_3\).
Fix one of them. By 6, an element \(dP_\sigma\) has order \(12\) exactly when the product of the three diagonal entries of \(d\) is \(i\) or \(-i\). For each prescribed product in \(\mu_4\), the first two diagonal entries may be chosen arbitrarily in \(4^2=16\) ways, and then the third is uniquely determined. Thus there are \[ 2\cdot16=32 \] valid diagonal matrices for each fixed \(3\)-cycle. Multiplying by the two choices of \(3\)-cycle gives \[ 2\cdot32=64 \] elements of order \(12\).