Prove that every group of order $4$ is abelian

Consider a group $G$ of order $4$. Suppose, towards a contradiction, that $G$ is not abelian. Then there must exist some distinct non-identity elements $x,y\in G$ such that $xy\ne yx$. But notice that:

• $xy≠e$ and $yx≠e$ (since $x$ and $y$ don’t commute, so $y≠x^{-1}$)
• $xy≠x$ and $yx≠x$ (since by hypothesis $y≠e$)
• $xy≠y$ and $yx≠y$ (since by hypothesis $x≠e$)

Thus, it follows that $e,x,y,xy,yx$ are $5$ distinct elements that are all in $G$. But this contradicts the fact that $G$ is of order $4$. Thus, $G$ must be abelian, as desired.

Let $G$ be a group of order $4$. If $G$ is cyclic, we're done. If not, $x^2 = e$ for all $x \in G$, which implies $(xy)(xy) = e$ for all $x, y \in G$. Multiply on the right by $yx$ to get $xy = yx$.

HINT: There are only two groups of order $4$. One of them has an element of order $4$; the other one doesn’t.