Let $(G, .)$ a group. Suppose $G = \{1, x, y, z\}$ is of order 4. Show that $G$ is abelian.
Am I right to say that, as $G$ is of order $4$, then by Lagrange's Theorem, there exists $g \in G$ such that $ \langle g\rangle = G$. From there it becomes clear that $G$ is abelian. Am I on the right track?
EDIT: There were two exercises before asking if $xy = x$ then $y = 1$ and if $xy \not = 1$ then $yx \not = 1$. Do I have to use those?
If $g\in G$ has order $4$, then $G=\langle g\rangle$;
otherwise, since the order of $g$ divides $4$, we get $g^2=1$, for every $g\in G$.
Thus $$xy=x1y=x((xy)(xy))y=(xx)(yx)(yy)=1(yx)1=yx.$$