Linear algebra and regular orbits

Question

If $A$ is an $n\times n$ matrix over a field, and $A^{k} = I$, with $k$ the least positive integer such that this occurs, then must there be some vector $v$ such that $\{v,Av,A^{2}v,\dots,A^{k-1}v\}$ has $k$ distinct elements in it? In other words:

Must every matrix of finite multiplicative order have a regular orbit?

If A has prime power order, $k = p^{m}$, then $A^{p^{m-1}}-I$ is nonzero, so its kernel is proper, and everything outside of that kernel is a vector in a regular orbit. Over a finite field of size $q$, the index of a proper subspace is at least $q$, so we can even just choose (on average) $q$ random vectors to find one in a regular orbit. Over an infinite field, the same idea roughly says any random vector should work, as long as one can make some sort of "uniformly" distributed choice.

If $A$ has order a product of two prime powers, then I am assured this is true by a (special case) of an exercise in Isaacs's Finite Group Theory. I cannot imagine an argument that does not work for arbitrary orders $k$, but I also cannot find a convincing proof even for the product of two prime powers. The sum of vectors in regular orbits of the $p$ - parts of $A$ need not themselves be in regular orbits of $A$. Every matrix (over a finite field) I've tried has a regular orbit.

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Assuming this is easy, how does one handle the case where $A$ is an automorphism of a finite group $G$, and the order of $A$ is a product of two prime powers? In other words:

Prove every automorphism of order $p^{a}q^{b}$ of a finite group has a regular orbit.

Assuming the first question's answer is "yes", then what goes wrong for arbitrary orders? Isaacs's book gives an example where the general automorphism can fail to have a regular orbit, but it is impossible to compare this until I have at least some idea of why the two-prime case does work.

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A related version of this question is: regular orbits are quite important in permutation and (finite) matrix groups and are a standard technique in several important (solved and unsolved) problems in modular representation theory.

Is there sort of a gentle introduction that puts these techniques in context?

For any individual paper is clear that what they say works, but my picture of this area is incredibly disjointed and I suspect that is not true for everyone. For instance Khukhro has an excellent book on automorphisms of p-groups with few fixed points, and many finite group theory texts have chapters on fixed-point-free automorphisms and the consequences for the group structure of the group being acted upon. However, I haven't found any "textbook" exposition of regular orbits yet.

Answer 1

For your first question, I presume you also wish to insist that $k$ be the least integer such that $A^k=I$. The matrix $A$ is then similar over your field to a direct sum $B_1,\ldots,B_m$ of companion matrices of (over your field $F$) factors of $X^k-1$, say $f_1,\ldots,f_m$. Then $F^n$ decomposes as a direct sum of subspaces where $A$ acts cyclically with generator $v_i$ annihilated by $f_i(A)$.

Let $v=v_1+\cdots +v_m$. Then for a polynomial $g$, $g(A)v=0$ if and only if $g(A)v_i=0$ for all $i$ if and only if $f_i\mid g$ for all $i$. Hence $F\mid g$ where $F=f_1\cdots f_m$. But then $F(A)u=0$ for all $u$ (so that $F$ is the minimum polynomial of $A$). If $A^l=I$ where $l < k$ then $F\mid(X^l-1)$ and then $A^l-I=0$, contrary to hypothesis. So yes, $A$ has a regular orbit.

Answer 2

Thanks to Marty Isaacs for reminding me the exercise was supposed to be easy and for giving a reference to my larger question. I'll post it here, since I think some people are following it.

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If the order of A is p^aq^b then the subgroup generated by A has two minimal subgroups, generated by P=A^pa−1qb and Q=A^paqb−1. If the orbit of g under A is not regular, then the stabilizer is a non-identity subgroup of A, so it contains either P or Q. Hence either P or Q centralize g. Hence g centralizes either P or Q (I am always amazed at the power of noticing "centralize" is symmetric), so g is in the union C_G(P) ∪ C_G(Q). Both of these subgroups are proper subgroups of G, since A acts faithfully on G itself. However, G is not the union of two proper subgroups, so there is some g in G − ( C_G(P) ∪ C_G(Q) ), and such a g represents a regular orbit.

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An investigation of which groups G must have regular orbits is in:

Horoševskiĭ, M. V. "Automorphisms of finite groups." Mat. Sb. (N.S.) 93(135) (1974), 576–587, 630. ^{(Math. USSR Sbornik 22 (1974) 4, 584–594)} MR347979 DOI: 10.1070/SM1974v022n04ABEH001707

and in particular proves that every automorphism of a nilpotent group or a semisimple (that is, Fitting-free) group has a regular orbit. The paper has exercise 3A.8 as a remark after corollary 1 (page 592 in the English translation), and corollary 3.3 as theorem 2. Its lemma 4 fixes my difficulties with dealing with the orbits prime by prime (don't look at the p-parts of A where the obvious statement has obvious counterexamples, look at non-faithful orbits instead).

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I could not generalize Robin Chapman's argument to finite abelian groups, since one no longer has that finite Z/nZ[t] modules are direct sums of cyclic modules (for instance, Z[t]/(4,tt-1) has the ideal (2,t+1)/(4,tt-1) of type C₂ × C₄ with t=A acting as the matrix [1,2;0,1]. This module is non-cyclic and indecomposable. Of course t has a regular orbit, but I could not simply choose a "generator". Marty Isaacs has shown me how to use Horoševskiĭ's argument to reduce to the case where G is indecomposable, where presumably it is easier than I think.

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In the other direction, keeping G elementary abelian but letting A be an entire group of automorphisms, one is still quite interested in whether there is a regular orbit. I found this article helpful for getting an idea of how this works:

Fleischmann, Peter. "Finite groups with regular orbits on vector spaces." J. Algebra 103 (1986), no. 1, 211–215. MR860700 DOI: 10.1016/0021-8693(86)90180-8.

In particular, nilpotent groups tend to have regular orbits except when p=2 is involved (either in A or G), and the specific problems with p=2 are addressed. Its methods for abelian groups A give an alternative view of Robin Chapman's answer (basically the paper shows that you can reduce to the algebraically closed/absolutely irreducible case, and then the Bi are all 1x1, and the matrix A is diagonal).