Irreducible cyclic groups of classical groups

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<X 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)<H;
  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).

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One thought on “Irreducible cyclic groups of classical groups

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  1. 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 🙂

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