Let \(\FF_p\) be the finite field with \(p\) elements, where \(p\) is a prime. Show that the centers of \(\GL_n(\FF_p)\) and \(\SL_n(\FF_p)\) consist only of scalar matrices.
- Show that the scalars \(\zeta\) that appear in scalar matrices \(Z(\SL_n(\FF_p))\) are roots of unity in \(\FF_p\), i.e. \(\zeta^p = 1\).
The scalars in \(Z(\SL_n(\FF_p))\) are the \(n\)th roots of unity in \(\FF_p\), \(\zeta^n=1\). The condition \(\zeta^p=1\) in the statement holds only for \(\zeta=1\), since \(\zeta^p=\zeta\) for every \(\zeta\in\FF_p\).
Solution
For \(i\neq j\) let \(e_{ij}\) be the matrix unit and \(T_{ij}=I_n+e_{ij}\), a transvection with \(\det T_{ij}=1\), so \(T_{ij}\in\SL_n(\FF_p)\subseteq\GL_n(\FF_p)\).
1. A matrix \(A=(a_{k\ell})\) commuting with every \(T_{ij}\) is scalar.
\(AT_{ij}=T_{ij}A\) gives \(Ae_{ij}=e_{ij}A\). The \((k,\ell)\) entries are \(a_{ki}\delta_{j\ell}\) and \(\delta_{ki}a_{j\ell}\). Taking \(\ell=j\) and \(k\neq i\) gives \(a_{ki}=0\); taking \(k=i\), \(\ell=j\) gives \(a_{ii}=a_{jj}\). As \(i\neq j\) vary, \(A\) is diagonal with equal diagonal entries.
2. \(Z(\GL_n(\FF_p))=\{\lambda I_n:\lambda\in\FF_p^\times\}\).
A central element commutes with every \(T_{ij}\), so it is scalar by step 1, and every invertible scalar matrix is central.
3. \(Z(\SL_n(\FF_p))=\{\zeta I_n:\zeta\in\FF_p,\ \zeta^n=1\}\cong\ZZ/\gcd(n,p-1)\ZZ\).
A central element of \(\SL_n(\FF_p)\) commutes with every \(T_{ij}\in\SL_n(\FF_p)\), so it is scalar by step 1; the scalar matrix \(\zeta I_n\) lies in \(\SL_n(\FF_p)\) exactly when \(\det(\zeta I_n)=\zeta^n=1\), and it is then central. The solutions of \(\zeta^n=1\) in the cyclic group \(\FF_p^\times\) of order \(p-1\) form its subgroup of order \(\gcd(n,p-1)\).