Non-Abelian Group has Order Greater than 4/Proof 1

Proof
Let $\left({G, \circ}\right)$ be a non-abelian group whose identity is $e$.

For a group $\left({G, \circ}\right)$ to be non-abelian, we require:
 * $\exists x, y \in G: x \circ y \ne y \circ x$

Suppose $x \circ y \in \left\{ {x, y, e}\right\}$.


 * $x \circ y = e \implies y \circ x = e$

and $\left({G, \circ}\right)$ is abelian.

, suppose $x \circ y = x$.

and again, $\left({G, \circ}\right)$ is abelian.

Similarly for $x \circ y = y$.

Again, the same applies if $y \circ x \in \left\{ {x, y, e}\right\}$.

So, if $x \circ y \ne y \circ x$, then;
 * $x \circ y$ and $y \circ x$ must be different elements
 * $x \circ y$ and $y \circ x$ must both different from $e, x$ and $y$.

Thus, in a non-abelian group, there needs to be at least $5$ elements:


 * $e, x, y, x \circ y, y \circ x$