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Let $n>1$ be a positive integer and let $R$ be an order in an imaginary quadratic field with discriminant prime to $n$. Let $A=R/nR$ and let $\lbrace 1, \alpha \rbrace$ be a $\mathbb{Z}/n\mathbb{Z}$ basis for $A$.

We know that for each prime $p|n$, $A/pA$ is isomorphic to either $\mathbb{F}_p \times \mathbb{F}_p$ or $\mathbb{F}_{p^2}$. In the latter, we say A is non-split at $p$. For each $a \in A^{\times}$, the multiplication by $a$ map induces a $\mathbb{Z}/n\mathbb{Z}$ linear bijection. So with respect to the $\mathbb{Z}/n\mathbb{Z}$ basis that we chose, $A^{\times}$ embeds into $GL_2(\mathbb{Z}/n\mathbb{Z})$.

A non-split Cartan subroup of $GL_2(\mathbb{Z}/n\mathbb{Z})$ is a subgroup that arises as an image of such $A^{\times}$ in this way and for which $A$ is non-split at all $p|n$.

I would like to know how one can use Hensel's Lemma to prove that all non-split Cartan subroups of $GL_2(\mathbb{Z}/n\mathbb{Z})$ are conjugate.

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  • $\begingroup$ Any particular reason you want a proof using Hensel's Lemma? $\endgroup$
    – Jeremy Rickard
    Commented Apr 26, 2015 at 17:22
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    $\begingroup$ In the paper that I am reading it says that it follows from Hensel's Lemma, but I could not see why. $\endgroup$
    – Jake Haider
    Commented Apr 26, 2015 at 17:23
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    $\begingroup$ The conjugacy class of the image is independent of the choice of basis of $A$ by definition, so it suffices to show that any two such $A$'s are isomorphic as $\mathbf{Z}/(n)$-algebras. Writing $n=\prod_{p|n} p^{e_p}$, we have $A=\prod_{p|n} A_p$ with primary parts $A_p$ that are $\mathbf{Z}/(p^{e_p})$-algebras, and explicitly $A_p = R_p/(p^{e_p})$ where the $p$-adic completion $R_p$ of $R$ is the valuation ring of a quadratic unmarried (!) extension $K_p$ of $\mathbf{Q}_p$. Such $K_p$ is unique up to isomorphism by Hensel's Lemma, so likewise for $R_p$ and hence $A_p$. QED $\endgroup$
    – grghxy
    Commented Apr 26, 2015 at 19:23
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    $\begingroup$ Dear @grghxy, By unmarried, you mean unramified? :) $\endgroup$
    – Keenan Kidwell
    Commented Apr 26, 2015 at 19:47
  • $\begingroup$ @KeenanKidwell: Hmm, auto-correct strikes again. $\endgroup$
    – grghxy
    Commented Apr 26, 2015 at 20:18

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