I was going through a Problem in an existing Thread: Find the Least Integer $k$ such that $B^k=I$
In the Approach given by OP and Omno we get: $$B^{63}=I$$
But How can we Justify that $63$ is Least?
My try:
I tried to check all powers of $B$ till $62$
We have $B \ne I$
Let $B^2=I$ Then $AB^2=BA$ $\implies$ $A=BA$ $\implies$ $B=I$ a Contradiction , Hence $$B^2\ne I$$
Also:
$$B^{63}=I$$ $$B^{62}=B^{-1}\ne I$$ $\implies$ $$B^{31}\ne I$$
So i could easily prove that
$B,B^2,B^{31}, B^{62}\ne I$
But now how to prove other powers of $B$ less than $63$ arealso not $I$?
The least $k$ can be smaller than $63$. E.g. when $$ A=\pmatrix{0&I_3\\ I_3&0}, \ B=\pmatrix{C&0\\ 0&C^2}, \ C=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}, $$ we have $A^2=A^6=I$ and $AB^2=\pmatrix{0&C\\ C^2&0}=BA$ but $B^3=I\ne B$.
(In the above counterexample, $A$ is annihilated by $x^2-1$. Perhaps the least $k$ is $63$ when the minimal polynomial of $A$ is $x^6-1$, but I don't know whether this hypothesis is true or not.)