Knowing that $A$ and $B$ are two invertible nxn matrices ($A≠I$, $B≠I$) such that: $A^7=I$ and $ABA^{-1}=B^2$ Show that there is an integer $k>0$ such that $B^k=I$ and determine the lowest value for $k$.
I realize that performing $(ABA^{-1})^x)$ we get $AB^xA^{-1}$. I tried to form an argument replacing the $B^x$ to a something like $A^zB^y(A^{-1})^z$ and when $z$ is 7 or a multiple, we would be able to get away with $A$ using $A^7=I$. I think this idea of performing $(ABA^{-1})^x)$ is right and very useful, but I dont't know if the repetitive ideia of replacing $B^x$ is the best to prove the first part of the exercise. And thinking this way, I don't know if I can formalize a solid argument to find the lowest $k$.
Since $A^7 = I$, $$\begin{align} B = & A^7BA^{-7}\\ = & A^6(ABA^{-1})A^{-6} = A^6B^2A^{-6}\\ = & A^5(AB^2A^{-1})A^{-5} = A^5(ABA^{-1})^2A^{-5} = A^5(B^2)^2A^{-5} = A^5B^4A^{-5}\\ \vdots & \\ = & B^{2^7} = B^{128} \end{align}$$ Since $B$ is invertible, multiply both sides by $B^{-1}$ gives us $B^{127} = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I \land B^{127} = I \implies B^{\gcd(127,m)} = I$$ If $m < 127$, the fact $127$ is a primes forces $\gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B \ne I$. As a result, $m \ge 127 \implies m = 127$.