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.