Let \(\mathbb F_2\) be the field with two elements.
What is the order of \(\text{GL}_3(\mathbb F_2)\)?
Use the fact that \(\text{GL}_3(\mathbb F_2)\) is a simple group (which you should not prove) to find the number of elements of order 7 in \(\text{GL}_3(\mathbb F_2)\).
Solution
1. The order of \(\operatorname{GL}_3(\mathbb F_2)\) is\[ (2^3-1)(2^3-2)(2^3-2^2)=7\cdot6\cdot4=168. \]
The first column of an invertible \(3\times3\) matrix may be any nonzero vector of \(\mathbb F_2^3\), giving \(8-1=7\) choices. The second column may be any vector outside the span of the first, giving \(8-2=6\) choices. The third may be any vector outside the \(2\)-dimensional span of the first two, giving \(8-4=4\) choices. Multiplying gives \(168\).
2. The number \(n_7\) of Sylow \(7\)-subgroups is \(8\).
Since \[ 168=2^3\cdot3\cdot7, \] the Sylow theorems give \[ n_7\equiv1\pmod7, \qquad n_7\mid24. \] The only divisors of \(24\) congruent to \(1\) modulo \(7\) are \(1\) and \(8\). If \(n_7=1\), the unique Sylow \(7\)-subgroup would be normal. Because \(\operatorname{GL}_3(\mathbb F_2)\) is simple and has order greater than \(7\), this is impossible. Hence \(n_7=8\).
3. There are exactly \(48\) elements of order \(7\) in \(\operatorname{GL}_3(\mathbb F_2)\).
Every Sylow \(7\)-subgroup has order \(7\), hence is cyclic and has exactly \(6\) nonidentity elements, all of order \(7\). Distinct subgroups of order \(7\) intersect only in the identity, since their intersection is a subgroup whose order divides \(7\). Thus the \(8\) Sylow \(7\)-subgroups contribute disjoint sets of \(6\) elements of order \(7\), giving \[ 8\cdot6=48. \]