Efficiently finding the not-quite-order of an element of $\text{Aut}((\mathbb{Z}/p\mathbb{Z})^2)$

Question

The motivation for this question is a Project Euler problem, but since I've heard posting solutions to those problems is discouraged, and an answer to this question makes the problem I'm working on very tractable, I'm leaving out specifics unless asked.

Question

I'm not sure if there's an actual term for what I'm calling the "not-quite-order" of a group element $g$, but what I mean is this:

Let a group $G$ act on a set $X$, with $x\in X$. The not-quite-order of $g$ with respect to $x$ is the smallest $k\in\mathbb{N}$ so that $g^k$ is in the stabilizer of $x$.

This question is interested in the family of groups $X=(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/p\mathbb{Z})$, where $p$ is is a prime other than 2, 3, or 7.

The function $$\varphi((a,b))=(a+7b,\thinspace a+b)$$ is an automorphism on $X$. Letting $\text{Aut}(X)$ act on $X$ in the obvious way, can I find the not-quite-order of $\varphi$ with respect to $(1,0)$ with less work and computational pain than the approach described below?

Work so far

I'll start by saying my current approach is to just naïvely compose $\varphi$ with itself until I find a $\varphi^k$ that fixes $(1,0)$. So it seems like I can only improve from here.

Any not-quite-order of $g$ must divide the order of $g$, and therefore must also divide $\lvert\text{Aut}(G)\rvert$. The largest $k$ can go is $p^2$, when $\text{Aut}(X)$ acts transitively. There are $(p^2-1)(p^2-p)$ automorphisms of $G$,* so between these three facts, it's appealing to think about only checking $\varphi^k$ if $k$ is a divisor of $(p^2-1)(p^2-p)$ less than $p^2$, and skipping numbers in between. All this, though, skips past the distinction between not-quite-orders and the order of $g$. This, though, still feels like guessing -- just guessing with improved efficiency -- rather than having a good strategy for finding the answer for very many, very large $p$. Are there ways to sharpen the upper bound on the not-quite-order of $\varphi$? Or ways we can exclude more divisors from this approach?

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$^*$ At least, I think there are. Say $\phi$ is an automorphism. Every element of $G$ other than the identity is of order $p$ and generates a cyclic group, so $\phi$ takes $(1,0)$ and maps it to one of these $p^2-1$ elements. Now $(0,1) \not\in \langle(1,0)\rangle$, so $\phi$ must map $(0,1)$ to an element not in $(\langle \phi(1,0)\rangle)$. There are $p^2-p$ ways to do that. Mapping the generators fully determines the automorphism and gives us our count of $(p^2-1)(p^2-p)$. This all sounds convincing to me, but the next time I get a counting argument right without any stumbles will be the first.

Answer 1

Too long for comment; still working on it, but these observations ought to be useful already.

1. An easy way to confirm that your $G$ has order $(p^2-1)(p^2-p)$ is to note that $X$ is the vector space of dimension $2$ over the field with $p$ elements, so $G$ is $\mathsf{GL}(2,p)$, the general linear group of degree $2$. The basis $(1,0)$, $(0,1)$ must map to a linearly independent set, and then your counting is clear: if $\varphi$ is an automorphism, then $(1,0)$ can be mapped to any nonzero element, and $(0,1)$ to any element outside the one-dimensional subspace generated bu $\varphi(1,0)$. See more generally this question about $\mathsf{GL}(n,p)$ and $\mathsf{SL}(n,p)$.

2. You can do better in terms of the orders of elements of $G$. This question asks about the order of elements in $\mathsf{GL}(2,q)$, with $q$ a power of a prime. For $q=p$, if the element is diagonalizable then it has order dividing $p-1$; if it has a repeated eigenvalue but is not diagonalizable, then the order is a multiple of $p$ that divides $p(p-1)$; and if it has no eigenvalues, then the order is a divisor of $p^2-1$ (and may be exactly $p^2-1$).

The element you have has matrix with respect to the standard basis given by $$\left(\begin{array}{cc} 1 & 7 \\ 1 & 1 \end{array}\right)$$ which has characteristic polynomial $(1-t)^2 - 7 = t^2 - 2t - 6$. The discriminant is $28=4\times 7$, and since $p$ is not $2$ or $7$, that means the discriminant is nonzero. So the matrix definitely does not have a repeated eigenvalue. (The reason we exclude $p=3$ is because in that case $\varphi$ is not invertible...)

If $7$ is a square modulo $p$ (using quadratic reciprocity, this happens for $p\gt 7$ exactly if $p\equiv 1$, $3$, $9$, $19$, $25$, or $27 \pmod{28}$), then the matrix is diagonalizable and has order $p-1$. If $7$ is not a square modulo $p$ (again, from quadratic reciprocity we know this means $p\equiv 5$, $11$, $13$, $15$, $17$, or $23\pmod{28}$), then the order divides $p^2-1$.

The eigenvalues are $1+\sqrt{7}$ and $1-\sqrt{7}$, with $\sqrt{7}$ the modular square root modulo $p$ when there is one; and an adjoined element when there is no square root modulo $p$.

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Now, the stabilizer of $(1,0)$ is precisely the set of linear transformations of the form $$\left(\begin{array}{cc} 1 & a\\ 0 & b \end{array}\right)$$ with $b\neq 0$. The eigenvalues are $1$ and $b$.

If the eigenvalues of $\phi$ are $\lambda_1$ and $\lambda_2$, then the eigenvalues of $\phi^k$ are $\lambda_1^k$ and $\lambda_2^k$. So you would need one of $(1+\sqrt{7})^k$ and $(1-\sqrt{7})^k$ to be equal to $1$.