If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$. Let $(G,*)$ be a group. If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.
$a^2*b=b*a \implies ab=a^{-1}*b*a$.
I can't proceed further.
You just have to combine your two equations!
If $a^4 = e$ and $a^2b = ba$, then
$$
b = a^4b = a^2(a^2b) = a^2(ba) = (a^2b)a = (ba)a = ba^2
$$
and therefore $a^2 = e$. But if $a^2 = e$, then $b = a^2b = ba$, so $a=e$.
Hope that helps,
Proof:
As a⁴=e. Then
b=a⁴b
b=(a²)²b
since a²b=ba
b=ba²
Since G is a group
b=a²b
Since
a²b=ba
b=ba=&gt;a=e.