Singer Cycle

From Finite Field

Definition

Let $F={\mathbb{F}}_q$ where $q=p^f$, and let $\rho$ be a generator of $F^{\times }$. The transformation on ${\mathbb{F}}_q^{+}\cong {\mathbb{F}}_p^f$ induced by $\rho$ is called a Singer cycle.

Proposition

Define $G:=F^{+}:⟨\rho ⟩={\mathbb{Z}}_p^f:{\mathbb{Z}}_{p^f-1}⩽\mathrm{Hol}(F^{+})={\mathbb{Z}}_p^f:{\mathrm{GL}}_f(p)$, where for any $a\in F^{+}$ and $b\in F^{\times }$, $a^b:=a\cdot b$. Then

• $G$ is sharply $2$-transitive on $F^{+}$; and
• $\mathrm{Aut}(G)=G:⟨\tau ⟩$ where $\tau$ is the Frobenius automorphism acting on ${\mathbb{Z}}_{p^f-1}$.

\begin{proof} i) For any distinct ordered pairs $(w_1,w_2)$ and $(v_1,v_2)$ in $F^{+}\times F^{+}$, there is $ab\in G$ such that $(w_i+a)\cdot b=v_i$. Then $(v_1-v_2)=(w_1-w_2)\cdot b$ yields that $b$ is unique and so for $a$.

ii) See here.

\end{proof}

From Linear Group

Definition

A Singer cycle of ${\mathrm{GL}}_n(q)$ is an element of order $q^n-1$.

Kantor - 1980 - Linear groups containing a singer cycle.pdf

If $G$ is a subgroup of $\mathrm{GL}(n,q)$ containing a singer cycle, then $G⊳\mathrm{GL}(n/s,q^s)$ for some $s$, embedded naturally in ${\mathrm{GL}}_n(q)$.