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I am trying to show that all cyclic subgroups of order $p^2-1$ are conjugate to one another in $GL_2(\mathbb{F}_p)$ but I am having a lot of difficulty. I have seen claims that each element in such a subgroup is uniquely determined by their characteristic polynomial, but given two cyclic groups of order $p^2-1$, how do I know the same conjugation will work for all elements?

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The eigenvalues of $A\in G=\text{GL}_2(\Bbb F_p)$ satisfy a quadratic, so lie in the cyclic group $\Bbb F_{p^2}^\times$. If $A$ has order $p^2-1$ its eigenvalues are outside $\Bbb F_p$ and so are $\alpha$ and $\alpha^p$ for some $\alpha$ which must generate $\Bbb F_{p^2}^\times$. If we have another $B$ of order $p^2-1$ its eigenvalues are $\beta$ and $\beta^p$ for some $\beta$. Then $\beta=\alpha^r$ for some $r$, and then $B$ and $A^r$ have the same distinct eigenvalues. Then $A^r$ and $B$ have the same rational canonical form, so are conjugate in $G$. Thus the groups they generate are conjugate.

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Since you are working with cyclic groups. You only need a conjugation on generators. Say, if $gxg^{-1} = y$ for some $g$, then for all $n\in \mathbb{Z}$ $$gx^{n} g^{-1} = (gxg^{-1})^n = y^n.$$ Thus $g\langle x \rangle g^{-1} = \langle y \rangle.$

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  • $\begingroup$ This is the way to address the last sentence of the question. But then you need to show that each such cyclic subgroup contains a generator with the same characteristic polynomial? $\endgroup$ – Jyrki Lahtonen Jun 29 '17 at 5:02
  • $\begingroup$ Yes, this is what I want to show. I want to do a pigeonhole argument where I argue the number of generators exceed the number of conjugacy classes. A generator necessarily has two eigenvalues in a field extension $\mathbb{F}_{p^2}$, and so there are $(p^2-p)/2$ possibly conjugacy classes for a generator and $\varphi(p^2-1)$ generators, but I cannot show that $\varphi(p^2-1) > (p^2-p)/2$. $\endgroup$ – Rdrr Jun 29 '17 at 5:34

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