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][1].

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


  [1]: https://mathoverflow.net/questions/327168/centralizers-of-cartan-subgroups
  [2]: http://pi.math.cornell.edu/~zywina/papers/Serre-Bound.pdf