Group generated by $x,y$. With relations $x^3=y^2=(xy)^2=1$. 
Let $G$ be a group generated by $x,y$ with the relations $x^3=y^2=(xy)^2=1$. Then show that the order of $G$ is 6.

My attempt: So writing down the elements of $G$ we have $\{1,x,x^2,y,\}$. Other elements include $\{xy, xy^2, x^2y\}$ it seems I am counting more than $6$. Are some of these equal? how do I prove that?

Since $y^2=1$ so $xy^2=x$. So only one of them counts. Thus $6$ elements. The given relation is basically resembles to the structure of $D_3$, the dihedral group of order $6.$
