What are the cyclic subgroups of $GL_2(\mathbb{Z}/p\mathbb{Z})$, the general linear group over the finite field of order $p$, where $p$ is prime?

Obviously, each cyclic subgroup is generated by some element $g \in GL_2(\mathbb{Z}/p\mathbb{Z})$. What are the possible orders of these cyclic subgroups, and how do I explicitly find a $g$ that generates each (i.e., explicitly find the elements of the $2\times 2$ matrix that corresponds to $g$)?

I can see that their order must divide $(p+1)p(p-1)^2$, but I can't immediately see exactly which orders are possible or how to construct them. I can see how to generate a cyclic subgroup of order $p-1$ (use the matrix $\left(\begin{smallmatrix}x &0\\0 &x\end{smallmatrix}\right)$, where $x \in (\mathbb{Z}/p\mathbb{Z})^*$ has order $p-1$) or of order $(p-1)p$ (use $\left(\begin{smallmatrix}x &1\\0 &x\end{smallmatrix}\right)$), from which it follows how to find a group whose order is any divisor of $(p-1)p$. However I'm having trouble seeing whether there's a cyclic subgroup of order $p+1$ (for example) nor how to construct it.


You can think of $\mathbb{F}_{p^2}$ as a 2-dimensional vector space over $\mathbb{F}_p$. Multiplication by any non-zero element of $\mathbb{F}_{p^2}$ is then an $\mathbb{F}_p$-linear map on $\mathbb{F}_{p^2}$, and so in this way $\mathbb{F}_{p^2}^\times$ embeds into $GL_2(\mathbb{Z}/p\mathbb{Z})$. Since the multiplcative group of a finite field is cyclic, this tells you that there is an element of $GL_2(\mathbb{Z}/p\mathbb{Z})$ of order $p^2-1$.

In general, the conjugacy class of an element of $GL_2(\mathbb{Z}/p\mathbb{Z})$ is almost determined by its characteristic polynomial. It's a nice exercise to go through the possibilities for the characteristic polynomial (irreducible over $\mathbb{F}_p$, reducible with distinct roots, reducible with double root) and try to see what elements have such characteristic polys. The only case where you have two conjugacy classes with the same characteristic polynomial is if the polynomial has a double root.


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