Let $H$ denote the group of all $2\times 2$ invertible matrices over $\mathbb Z_5$ under usual matrix multiplication. Then the order of the matrix
$$ \begin{bmatrix} 2&3\\ 1&2\\ \end{bmatrix} $$ in $H$ is--?

It means we have to find order, i.e to which power we should raise this matrix to get an identity matrix. If we raise its power to $3$, we'll get the identity matrix. or can we consider this matrix as $SL_n(\mathbb Z_p)$ group and calculate its order by the formula $[(p^n-1)(p^n-p)...(p^n-p^{n-1})]/(p-1)$ which method is correct?

Is there any other direct formula to calculate that?

  • $\begingroup$ Your formula is for the order of the group, not for the order of an element of the group. $\endgroup$ – Tobias Kildetoft Mar 4 '16 at 9:19
  • $\begingroup$ Thank you for clarifying what i was assuming wrong. So correct way to find the order is to raise the power of matrix to get an identity matrix? $\endgroup$ – Sonali Jain Mar 4 '16 at 9:33
  • $\begingroup$ Yes, that is indeed the only way, and you have already done this it seems. $\endgroup$ – Tobias Kildetoft Mar 4 '16 at 9:34

Well, you may observe that the characteristic polynomial of your matrix $M$ is $X^2 - Tr(M)X + det(M)$, that is $X^2 -4X +1$, or $X^2 + X + 1$. It follows that $M^2 +M+1 = 0$, whence $M^3 -1=0$.

  • $\begingroup$ Than you for the answer. can you please explain how we got X^2 +X+1 after X^2- 4X+1 and then M^3 -1 =0? $\endgroup$ – Sonali Jain Mar 4 '16 at 9:51
  • $\begingroup$ @SonaliJain $\;-4=1\pmod5\;$ , and $\;M^3-1=(M-1)(M^2+M+1)\;$ $\endgroup$ – DonAntonio Mar 4 '16 at 10:00

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