Group $(G, \cdot )$ contains identity $e$ and two distinct non identity elements $x$ and $y$.
Given that $x \cdot y = y^2 \cdot x$
Prove $x$ does not commute with $y$.
To be clear this is not the dot product but just an unknown operation.
My initial instinct is simply to state that since they are distinct elements then that means they will be greater or smaller than each other, therefor squaring one of them will result in a different outcome than if you had squared the other; although I have a feeling this is not enough to prove the group is not commutative.
Best Answer
"Greater or smaller": there are groups on which it is impossible to impose a total order, so this can't be the right way.
Hint: you're going to have to show that $xy \not = y x$; equivalently $y^2 x \not = y x$. You can cancel some things out of that expression.