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Let $\mathcal O$ be an order in an imaginary quadratic field $K$. Let $n$ be a positive integer. The multiplicative group $(\mathcal O/n\mathcal O)^\times$ acts on the module $\mathcal O/n\mathcal O\cong \mathbb Z / n\mathbb Z\times \mathbb Z / n\mathbb Z$. We get thus an embedding of $(\mathcal O/n\mathcal O)^\times$ into $\operatorname{GL}_2(\mathbb Z / n\mathbb Z)$. Let $C_n$ be the image.

Is the centralizer of $C_n$ in $\operatorname{GL}_2(\mathbb Z / n\mathbb Z)$ equal to $C_n$ itself?

I am interested especially in the case when $n$ is a power of two.

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  • $\begingroup$ What is the difference to your previous question mathoverflow.net/questions/327168/… ? $\endgroup$ – Chris Wuthrich Apr 20 '19 at 23:12
  • $\begingroup$ @ChrisWuthrich, I still do not know how to prove this if $n$ is a power of two. $\endgroup$ – Shimrod Apr 21 '19 at 6:14

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