If $ABA = B$ and $BAB = A$ and $A$ is invertible, then $A^4 = I$ Let $A$ and $B$ be square matrices of the same order so that $ABA = B$ and $BAB = A$. If $A$ is invertible, prove that $A^4 = I$.
I already proved that $A^2=B^2$. How can I prove $A^4=I$?
 A: You have that $A$ and $B$ are square matrices of the same order with
$$ABA = B \tag{1}\label{eq1}$$
$$BAB = A \tag{2}\label{eq2}$$
You're asking that, if $A$ is invertible, to then prove that $A^4 = I$.
You've already proven that $A^2 = B^2$, but I am doing it here as well, just in case it's a different method than yours or if somebody reading this doesn't know how to do it. With \eqref{eq1}, multiply on the right by $B$, then use associativity of matrix multiplication, and finally \eqref{eq2}, to get
$$\begin{equation}\begin{aligned}
ABA(B) & = B(B) \\
A(BAB) & = B^2 \\
A(A) & = B^2 \\
A^2 & = B^2
\end{aligned}\end{equation}\tag{3}\label{eq3}$$
With \eqref{eq2}, left & right multiply by $B$, then use \eqref{eq3}, the associativity of matrix multiplication, and finally the invertibility of $A$ to multiply on the right by $A^{-1}$, to get
$$\begin{equation}\begin{aligned}
B(BAB)B & = B(A)B \\
(BB)A(BB) & = A \\
(AA)A(AA) & = A \\
(A^4)AA^{-1} & = AA^{-1} \\
A^4 & = I
\end{aligned}\end{equation}\tag{4}\label{eq4}$$
