One of the most useful things to know about the finite general linear group are the Singer cycles, what they do and what they are about. A subgroup of is *irreducible* if it fixes no proper nontrivial subspace of the vector space , in the natural action of on this vector space. Consider as a vector space over its subfield . Then the multiplicative group acts on the nonzero vectors of 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* , where divides , and it is defined by

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

Consider with odd. Let be the following form defined over :

Note that defines a non-degenerate Hermitian form on , as

and the map is the unique automorphism of of order 2. (We would get a degenerate form if were even as then ).

Now consider a primitive root of , and let . Then has order .

Now for all , we have

and hence is an isometry of . Finally, note that acts semiregularly on nonzero vectors (as it is a subgroup of a Singer cycle of and so each orbit on nonzero vectors has size , and the number of nonzero vectors in an i-dimensional subspace is . So acts irreducibly on as a -dimensional vector space over .

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

Consider with odd. Let be the following form defined over :

Note that is just the composition of the previous form for with the relative trace from to . Now defines a non-degenerate symmetric bilinear form on , as

.

Just as before, we consider a primitive root of , and let . Then has order and defines an isometry of . It is not difficult to see that acts irreducibly on , as a -dimensional vector space over .

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

Consider with odd. This time, we add a twist to the type of forms we have been using. Let be an element of such that . Let be the following form defined over :

Now defines a skew-symmetric bilinear form on , as

.

We then consider a primitive root of , and let . As per usual has order and defines an isometry of such that acts irreducibly on (thought of as a -dimensional vector space over ).

## 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 , , , , with , and is a maximal overgroup of an irreducible cyclic subgroup of X of maximal order. Then one of the following holds.

- H is an extension field type subgroup of X;
- and .
- and .
- .
- .
- .
- , even and ;
- ;
- .

(I have removed the cases , and for a simpler statement).

John, Just thought you might like to know that I’ve used this post at least a dozen times when I’ve needed to refresh my mind about Singery thingys. Thanks for posting it, and I hope all is good at UWA. nick 🙂