One of the most useful things to know about the finite general linear group $GL(d, q)$ are the Singer cycles, what they do and what they are about. A subgroup $G$ of $GL(d, q)$ is irreducible if it fixes no proper nontrivial subspace of the vector space $GF(q)^d$, in the natural action of $G$ on this vector space. Consider $GF(q^d)$ as a vector space over its subfield $GF(q)$. Then the multiplicative group $S:=GF(q^d)\backslash\{0\}$ acts on the nonzero vectors of $GF(q^d)$ by right multiplication, and this action is clearly regular. But what about similar elements in the classical groups?

An important map in the following will be the relative trace map $Tr_{q^d\to q^i}:GF(q^d)\to GF(q^i)$, where $i$ divides $d$, and it is defined by

$Tr_{q^d\to q^i}(x) = \sum_{j=0}^{d/i-1}x^{(q^i)^j}.$

## Singer-type elements of GU(d,q), d odd

Consider $GF(q^{2d})$ with $d$ odd. Let $B$ be the following form defined over $GF(q^2)$:

$B(x,y) := Tr_{q^{2d}\to q^2}(xy^{q^d}).$

Note that $B$ defines a non-degenerate Hermitian form on $GF(q^{2d})$, as

$B(y,x)=Tr_{q^{2d}\to q^2}(yx^{q^d})=Tr_{q^{2d}\to q^2}\left(\left(xy^{q^d}\right)^{q^d}\right)$

$=Tr_{q^{2d}\to q^2}(xy^{q^d})^{q^d}=B(x,y)^{q^d}$

and the map $z\mapsto z^{q^d}$ is the unique automorphism of $GF(q^{2d})$ of order 2. (We would get a degenerate form if $d$ were even as then $Tr_{q^{2d}\to q^2}(xy^{q^d}) =2 Tr_{q^d\to q^2}(xy^{q^d})$).

Now consider a primitive root $\xi$ of $GF(q^{2d})$, and let $\omega:=\xi^{q^d-1}$. Then $\omega$ has order $q^d+1$.

Now for all $i$, we have

$B(x\omega^i, y\omega^i)=Tr_{q^{2d}\to q^2}\left(x\omega^i\left(y\omega^i\right)^{q^d}\right)=Tr_{q^{2d}\to q^2}\left(xy^{q^d}\left(\omega^{q^d+1}\right)^{i}\right)=Tr_{q^{2d}\to q^2}\left(xy^{q^d}\right)=B(x,y)$

and hence $\omega$ is an isometry of $B$. Finally, note that $\langle \omega\rangle$ acts semiregularly on nonzero vectors (as it is a subgroup of a Singer cycle of $GL(d,q^2)$ and so each orbit on nonzero vectors has size $q^d+1$, and the number of nonzero vectors in an i-dimensional subspace is $q^{2i}-1$. So $\langle \omega\rangle$ acts irreducibly on $GF(q^{2d})$ as a $d$-dimensional vector space over $GF(q^2)$.

## Singer-type elements of GO-(2d,q), d odd

Consider $GF(q^{2d})$ with $d$ odd. Let $B$ be the following form defined over $GF(q)$:

$B(x,y) := Tr_{q^{2d}\to q}(xy^{q^d}).$

Note that $B$ is just the composition of the previous form for $GU(d,q)$ with the relative trace from $GF(q^2)$ to $GF(q)$. Now $B$ defines a non-degenerate symmetric bilinear form on $GF(q^{2d})$, as

$B(y,x)=Tr_{q^{2d}\to q}(yx^{q^d})=Tr_{q^{2d}\to q}\left(\left(xy^{q^d}\right)^{q^d}\right)$

$=Tr_{q^{2d}\to q}\left(xy^{q^d}\right)^{q^d}=Tr_{q^{2d}\to q}(xy^{q^d})=B(x,y)$.

Just as before, we consider a primitive root $\xi$ of $GF(q^{2d})$, and let $\omega:=\xi^{q^d-1}$. Then $\omega$ has order $q^d+1$ and $\omega$ defines an isometry of $B$. It is not difficult to see that $\langle \omega\rangle$ acts irreducibly on $GF(q^{2d})$, as a $2d$-dimensional vector space over $GF(q)$.

## Singer-type elements of Sp(2d,q)

Consider $GF(q^{2d})$ with $d$ odd. This time, we add a twist to the type of forms we have been using. Let $\gamma$ be an element of $GF(q^{2d})$ such that $\gamma^{q^d}+\gamma=0$. Let $B$ be the following form defined over $GF(q)$:

$B(x,y) := Tr_{q^{2d}\to q}(\gamma xy^{q^d})$

Now $B$ defines a skew-symmetric bilinear form on $GF(q^{2d})$, as

$B(y,x)=Tr_{q^{2d}\to q}(\gamma yx^{q^d})=Tr_{q^{2d}\to q}\left(\left(\gamma^{q^d}xy^{q^d}\right)^{q^d}\right)$

$=Tr_{q^{2d}\to q}\left(-\gamma xy^{q^d}\right)^{q^d}=-Tr_{q^{2d}\to q}(\gamma xy^{q^d})=-B(x,y)$.

We then consider a primitive root $\xi$ of $GF(q^{2d})$, and let $\omega:=\xi^{q^d-1}$. As per usual $\omega$ has order $q^d+1$ and $\omega$ defines an isometry of $B$ such that $\langle \omega\rangle$ acts irreducibly on $GF(q^{2d})$ (thought of as a $2d$-dimensional vector space over $GF(q)$).

## There are no others

By using Zsigmondy’s Theorem on primitive prime divisors, it is not difficult (but a little tedious) to show that the examples above are the only examples of irreducible cyclic subgroups of the classical groups.

## Overgroups of Singer-type elements

Áron Bereczky (2000) determined the maximal overgroups of Singer elements of the classical groups. Most of them preserve an extension field structure, but there are some interesting exceptions:

Theorem (Bereczky): Suppose X is one of the groups $SL(d,q)$, $SU(d,q)$, $Sp(d,q)$, $\Omega^-(d,q)$, with $d\ge 2$, and $H is a maximal overgroup of an irreducible cyclic subgroup of X of maximal order. Then one of the following holds.

1. H is an extension field type subgroup of X;
2. $d=2$ and $q\le 9$.
3. $d=3$ and $q\le 4$.
4. $(d,q)=(4,3)$.
5. $X=SU(5,2)$.
6. $X=Sp(8,2)$.
7. $X=Sp(d,q)$, $q$ even and $\Omega-(d,q);
8. $X=\Omega^-(6,2)$;
9. $X=\Omega^-(6,3)$.

(I have removed the cases $X=GU(d,q)$, $X = GO^-(d,q)$ and $X=SO^-(d,q)$ for a simpler statement).