Solution
= Solution
Suppose $x,y$ have order three but $x\ne y,y^2$. Their <cyclic subgroups> have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^iy^j$, $0\leq i,j<3$, are distinct. Indeed $x^iy^j=x^ky^\ell$ implies $x^{i-k}=y^{\ell-j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.