Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$. Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.
attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, and
$b^3a^3 = b^{-2}(ab^{-1})b^2$.
Denote the identity element of $G$ by $e_G$.
Fact: for all $a \in G$,
$|a| = |a^{-1}|$.
Let $|ba^{-1}| = m$ and $|b^3a^3| = p$.
To show $m \mid p$:
\begin{align*}
(b^3a^3)^p &= e_G \\
(b^{-2}(ab^{-1})b^2)^p &= e_G \\
b^{-2}(ab^{-1})^pb^2 &= e_G \\
b^{-2}((ba^{-1})^{-1})^pb^2 &= e_G \\
((ba^{-1})^{-1})^p &= e_G
\end{align*}
By fact, $|(ba^{-1})^{-1}| = m$.
Hence, $m \mid p$ as desired.
To show $p \mid m$:
\begin{align*}
(ba^{-1})^m &= e_G \\
((ab^{-1})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^m)^{-1} &= e_G
\end{align*}
Here I got a little bit confuse.
Does it true that in:
\begin{equation*}
((b^2(b^3a^3)b^{-2})^m)^{-1} = e_G
\end{equation*}
implies
$((b^2(b^3a^3)b^{-2})^m) = e_G$? (This is come from fact.) If yes, then
$|b^3a^3| = m$. Hence, $p \mid m$.
And thus, $m=p$.

You don't need to go through divisibility arguments. Since $ab^{-1}=(b^2)(b^3a^3)(b^{-2})$, $ab^{-1}$ and $b^3a^3$ are conjugate and have the same order. Using the fact about orders of inverses, the original claim follows.
