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Given matrices in $\Bbb{Z}_p$ equivalent to this form $$ \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} $$

What is the centralizer and its order?

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closed as off-topic by José Carlos Santos, Mohammad Riazi-Kermani, Arnaud Mortier, David Hill, Matthew Conroy Feb 9 '18 at 22:46

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Hint: A matrix $M$ will be an element of the centralizer if and only if it commutes with the matrix $$ \pmatrix{0&1\\-1&0} $$ since your set of matrices consists of those having the form $$ a \pmatrix{1&0\\0&1} + b \pmatrix{0&1\\-1&0} $$


Note that $$ \pmatrix{0&1\\-1&0}\pmatrix{m_{11} & m_{12}\\m_{21} & m_{22}} = \pmatrix{m_{21} & m_{22}\\-m_{11} & -m_{12}}\\ \pmatrix{m_{11} & m_{12}\\m_{21} & m_{22}}\pmatrix{0&1\\-1&0} = \pmatrix{-m_{12} & m_{11}\\-m_{22} & m_{21}} $$ These matrices are equal if and only if $m_{11} = m_{22}$ and $m_{12} = -m_{21}$.

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  • $\begingroup$ How can I obtain the order of centralizer? $\endgroup$ – User432477438 Feb 9 '18 at 15:28
  • $\begingroup$ Once you have figured out which elements commute with $\pmatrix{0&1\\-1&0}$, count them. $\endgroup$ – Omnomnomnom Feb 9 '18 at 15:41
  • $\begingroup$ Is there any easy method? Or obtain a formula that will count the number of centralizer? $\endgroup$ – User432477438 Feb 9 '18 at 15:44
  • $\begingroup$ See my latest edit. I'm just using the definitions. $\endgroup$ – Omnomnomnom Feb 9 '18 at 15:53
  • $\begingroup$ Therefore this form $$ \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} $$ is obtain? $\endgroup$ – User432477438 Feb 9 '18 at 15:56

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