Order of "one minus automorphism" This is something I am stuck on (it might well be trivial- in which case this is an embarassing question):
Let V be a dimension r vector space over Fp, the field with p prime elements (I also care about this when V is over Zn with n composite). Let t be a given automorphism of V of order q (prime different from p, although I am also interested in more general cases), meaning tq=1. edit: It is given that for any non-zero vector v in V, the elements of the set {ti v} together generate V, where i is between 1 and q. This implies that 1-t is also an automorphism of V. (thanks David!)
Question: What is the order of 1-t? When is it q?
The context is representations of commutator subgroups of knot groups onto vector spaces. Here t is induced by the deck transformations of the infinite cyclic cover.
 A: Perhaps the answer is just that the order is what it is, and that sometimes it's q. Do you have a compelling reason to believe that there is any more structure to your question than that?
Here's an example of the situation. $V$ could be the finite field $F$ with $p^r$ elements, and $q$ could be a prime divisor of $p^r-1$. Now $F^\times$ is cyclic of order $p^r-1$ so will contain elements of order $q$. Let $t$ be such an element. Multiplication by $t$ gives an endomorphism of $V$. The minimal polynomial of $t$ over the prime subfield will have degree dividing $r$, and if $q$ doesn't divide $p^s-1$ for any $s$ less than $r$ then it will have degree equal to $r$. So if $q\geq r$ and $q$ divides $p^r-1$ but not $p^s-1$ for any $s$ less than $r$, and such primes are easy to find (for example take $r$ to be 1---or let $r$ be prime; then for any given $p$ there is almost always a $q$) then the conditions of your question are satisfied (this "degree $r$" business is just making sure the vectors span). Now you're asking what the order of $1-t$ is, and it will be some divisor of $p^r-1$ and it might be $q$ and it might not be. 
For example let's choose everyone's favourite prime $p=691$, let $r-1$, let $q=23$ be the biggest divisor of $p-1$, and let's consider the 22 primitive 23rd roots of unity in $\mathbf{Z}/691\mathbf{Z}$. For each of these $t$, let's ask if 1-t is also a 23rd root of 1. The answer is a resounding "sometimes" (namely when $t=20$ or 672).
A: Your first condition implies that $r$ is the order of $p$ modulo $q$.
If the order of $1-t$ is $q$, then $p$ divides 
$$ \prod_{i=1}^{p-1} \prod_{j=1}^{p-1} (1 - \zeta^i - \zeta^j) \quad (*)$$
where $\zeta$ is a primitive $p$-th root of unity.
If these two conditions are satisfied, we can find $t$ with the required properties.
Proofs: 
First, suppose that $t$ obeys your first condition. Let $f$ be the characteristic polynomial of $t$. If $f$ factored as $gh$, and $v$ were in the kernel of $g(t)$, then $\mathrm{Span} (t^i v)$ has dimension at most $\deg g$, contradicting your hypothesis.
So $f$ is irreducible. In particular, $f$ divides $x^{p^r}-x$ and so $q | p^r-1$. Also, if $f$ divided $x^{p^s} - x$ for some $s<r$, then $f$ has a nontrivial factor of degree $\leq s$, contradicting that $f$ is irreducible. So $r$ is the order of $p$ modulo $q$.
Now, suppose that $1-t$ has order $q$. Let $\Phi_q(x) = (x^q-1)/(x-1)$. Then $f(x) | \Phi_q(x)$ and $f(1-x) | \Phi_q(x)$ as well. So $f$ divides $\mathrm{GCD}(\Phi_q(x), \Phi_q(1-x))$. In particular, the resultant of $\Phi_q(x)$ and $\Phi_q(1-x)$ is zero modulo $p$. The product $(*)$ is precisely that resultant.
Finally, we must reverse the argument. Suppose that $p$ divides $(*)$. So there is an irreducible polynomial $f$ dividing $\Phi_q(x)$ and $\Phi_q(1-x)$. Let $s = \deg f$. Let $V$ be the field $\mathbb{F}_{p^s}$ and let $t$ be the action of a root of $f$ on $V$. Then $t$ obeys condition 1. (If not, $\mathrm{Span} (t^i v)$ is a $t$-stable subspace of $V$, contradicting that $f$ is irreducible.) By the first paragraph of the proof, this shows that $s$ is the order of $p$ modulo $q$, so $s=r$ and we have constructed the desired $t$.
