Let \(F\) be a finite field of cardinality \(p^n\), where \(p\) is prime and \(n>0\), and let \[ G=GL_2(F). \]
- Prove that \[ |G|=(p^{2n}-1)(p^{2n}-p^n). \]
- Show that every Sylow-\(p\) subgroup of \(G\) is isomorphic to the additive group of \(F\).
Solution
Put \[ q\coloneqq\abs F=p^n. \]
1.\[ \abs{G}=(q^2-1)(q^2-q). \]
An invertible \(2\times2\) matrix is the same as an ordered basis of \(F^2\). Its first column can be any nonzero vector, giving \(q^2-1\) choices. Once the first column is fixed, its span contains exactly \(q\) vectors, so the second column has \(q^2-q\) choices outside that span. Multiplying gives the formula.
2. The largest power of \(p\) dividing \(\abs G\) is \(p^n=q\).
Step 1 gives \[ \abs G =(q^2-1)(q^2-q) =q(q-1)^2(q+1). \] Since \(q=p^n\), \[ q-1\equiv-1\pmod p, \qquad q+1\equiv1\pmod p, \] so neither \(q-1\) nor \(q+1\) is divisible by \(p\). Thus the full \(p\)-part of \(\abs G\) is \(q=p^n\).
3. The subgroup\[ U\coloneqq \left\{ \begin{pmatrix} 1&a\\ 0&1 \end{pmatrix} :a\in F \right\} \]is a Sylow-\(p\) subgroup of \(G\) and is isomorphic to \((F,+)\).
2. Thus \(U\) is Sylow-\(p\).
4. Every Sylow-\(p\) subgroup of \(G\) is isomorphic to \((F,+)\).
By the Sylow conjugacy theorem, every Sylow-\(p\) subgroup of \(G\) is conjugate to \(U\). Conjugate subgroups are isomorphic, and step 3 identifies \(U\) with \((F,+)\).
5. Q.E.D.
Step 1 proves part 1, and step 4 proves part 2.