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Let $K$ be an imaginary quadratic field and let $\mathcal O$ be its ring of integers. Suppose that $2$ is split in $\mathcal O$. Let $k$ be a positive integer. The multiplicative group $(\mathcal O/2^k\mathcal O)^\times$ acts on the additive group $\mathcal O/2^k\mathcal O$ by multiplication. Therefore there is an embedding $(\mathcal O/2^k\mathcal O)^\times \rightarrow \operatorname{GL}_2(\mathbb Z/ 2^k\mathbb Z)$. Let $C_k$ be the image. Is the centralizer of $C_k$ equal to $C_k$?

This follows my question here.

A similar assertion for odd primes is stated without proof in this paper, section 2.

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  • $\begingroup$ Do you mean to switch to $\mathbb Z/2^k/\mathbb Z$ after discussing $\mathcal O/2^k\mathcal O$? $\endgroup$
    – LSpice
    Commented Apr 8, 2019 at 11:32
  • $\begingroup$ @LSpice, What i meant was that $\mathcal O/ 2^k\mathcal O$ is isomorphic to $\mathbb Z/2^k\mathbb Z \times \mathbb Z/2^k\mathbb Z $ and that this gives us the embedding. $\endgroup$
    – Shimrod
    Commented Apr 8, 2019 at 11:37
  • $\begingroup$ Same answer as in the question this follows: the centraliser is contained in the normaliser, which is generated by $C$ and $(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$. the latter acting non-trivially on $C$ so $C$ is its own centraliser. Or is there something different for $p=2$? $\endgroup$
    – Chris Wuthrich
    Commented Apr 8, 2019 at 12:11
  • $\begingroup$ @ChrisWuthrich, I do not understand why should the matrix $\begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}$ generate the normaliser. Can you please explain that or direct me to some reference? $\endgroup$
    – Shimrod
    Commented Apr 8, 2019 at 12:39
  • $\begingroup$ You are in the split case, so you can choose the basis of $\mathcal{O}$ so that $C$ are just the diagonal matrices. Obviously that matrix is normalising and it is easy to check that the normaliser is $C\cup(\begin{smallmatrix}0&*\\*&0\end{smallmatrix})$. $\endgroup$
    – Chris Wuthrich
    Commented Apr 8, 2019 at 14:21

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