Non-Abelian Group has Order Greater than 4/Proof 2

Theorem
Let $\left({G, \circ}\right)$ be a non-abelian group.

Then the order of $\left({G, \circ}\right)$ is greater than $4$.

Proof
It follows from Trivial Group is Cyclic Group and Prime Group is Cyclic that groups of order less than $4$ are cyclic.

Therefore, by Cyclic Group is Abelian, all groups of order less than $4$ are abelian.

Let $G$ have order $4$.

From Order of Element Divides Order of Finite Group, every element of $G$ has order that divides $4$.

Therefore any element of $G$ has order $1$, $2$ or $4$.

Let $g \in G$ have order $4$.

Then $G$ is generated by $g$.

Therefore $G$ is cyclic, and therefore abelian by Cyclic Group is Abelian.

Let $G$ be such that it has no element of order $4$.

From Identity Only Group Element Order 1, all the elements apart from its identity must have order $2$.

That is, $G$ is a Boolean group.

By Boolean Group is Abelian, $G$ is abelian.

The result follows.