7

Cyclic subgroups of $\operatorname{GL}(n,q)$ of order $q^n - 1$ are called Singer cyclic subgroups. The following statement seems to be well-known:

Any two Singer cyclic subgroups of $\operatorname{GL}(n,q)$ are conjugate.

I don't know how to prove this in general. Sometimes, $q^n - 1$ is prime such that Sylow can be used (Example: $q=2, n=3$).

Some more background:

  • Elements of order $q^n - 1$ (and thus Singer cyclic subgroups) always exist: Let $\alpha$ be a primitive element of $\operatorname{GF}(q^n)$ and look at the $\operatorname{GF}(q)$-linear map $\operatorname{GF}(q^n) \to \operatorname{GF}(q^n)$ given by $x \mapsto \alpha x$.

  • $q^n - 1$ is the largest element order in $\operatorname{GL}(n,q)$, as by Cayley-Hamilton, the $\operatorname{GF}(q)$ vector space spanned by $I,A,A^2,A^3,\ldots$ has at most dimension $n$.

  • The elements of order $q^n - 1$ are not necessarily conjugate. For example, in $\operatorname{GL}(3,2)$ there are two conjugacy classes of elements of order $7$. However, the generated Singer cyclic subgroups are conjugate.

azimut
  • 22,696

2 Answers2

8

Here is a slightly expanded version of the proof mentioned by Dietrick Burde using the representation theory of cyclic groups.

A cyclic subgroup of order $q^n-1$ clearly clearly acts irreducibly on ${\mathbb F}_q^n$, so the minimal polynomial of a generator $g$ is a factor of $x^{q^n-1}-1$ of degree $n$ that is irreducible over ${\mathbb F}_q$.

So the subalgebra of $A_g \le M_n(q)$ generated by $g$ is isomorphic to the field of order $q^n$.

Since there is a unique such field up to isomorphism, for any other $g' \in {\rm GL}(n,q)$ of order $q^n-1$ generating a matrix algebra $A_{g'}$, there is an ${\mathbb F}$-isomorphism $\phi:A_g \to A_{g'}$, and the matrix defined by $\phi$ conjugates the set $\langle g \rangle = A_g \setminus \{ 0 \}$ to $\langle g' \rangle$.

Derek Holt
  • 90,008
6

The fact that all Singer "cycles" are conjugate follows from representation theory of cyclic groups. There is a detailed proof in the book "Finite groups" by B. Huppert, 1967, II. $§ 7$, page 187.

There is also a proof, that a Singer subgroup of $GL(n,q)$ is a maximal torus of the reductive group $GL(n,q)$, and that any two maximal tori are conjugate.

Reference: G. Hiss, Finite groups of Lie type, section $1.2$.

Dietrich Burde
  • 130,978
  • Many thanks for this expert answer and the references. Apparently, there is no "elementary" answer to my question. – azimut Mar 31 '16 at 12:58
  • Well, the representation theory proof could be considered as "elementary". – Dietrich Burde Mar 31 '16 at 13:02
  • That's why I put "elementary" into quotation marks :) – azimut Mar 31 '16 at 13:16
  • I just came back to this answer and checked the reference in Huppert. Satz 7.3 on page 187 clearly is about Singer cyclic subgroups. Maybe I'm blind, but I cannot find the statement under question, which is that any two of them are are conjugate. Dietrich Burde, would you mind to check the reference again? Thank you! – azimut Mar 13 '17 at 13:39
  • I do not have the book here right now. Many papers refer this fact to Huppert's book (in the German edition), so it "must" be there. If it is really not there, then you still have Derek's expanded answer (which you have accepted). – Dietrich Burde Mar 13 '17 at 13:50
  • Thank you for this quick response. Satz 7.3 basically says that in $\operatorname{GL}(n,q)$, Singer cycles do always exist, any singer cycle $S$ equals its centralizer, and has a cyclic quotient of order $n$ in its normalizer (representing a generator of $S$ as $\operatorname{GF}(q^n)\to\operatorname{GF}(q^n)$, $x\mapsto \alpha x$, a complement is given by the cyclic Galois group of order $n$ of the field extension $\operatorname{GF}(q^n)/\operatorname{GF}(q))$. There are more statements, aiming at the normalizers of cyclic subgroups and at the restriction/image of $S$ in SL, PGL and PSL. – azimut Mar 13 '17 at 14:09