Group of Order less than 6 is Abelian

Theorem
All groups with less than $6$ elements are abelian.

Proof

 * There is only one group of order $1$, and that is the Trivial Group which is trivially abelian.


 * All groups of prime order are cyclic and therefore abelian. So groups of order $2$, $3$ and $5$ are abelian.


 * We need to show that a non-abelian group has more than $4$ elements.

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

We know that $x \circ y = x \implies y = e$, and $x \circ y = y \implies x = e$, from the definition of the identity $e$.

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

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


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

Alternative Proof
The argument concerning groups of order $1$, $2$, $3$ and $5$ is as above.

Alternatively, let $G$ have order $4$.

From Order of Element Divides Order of Finite Group, every element of $G$ has order that divides $4$, so must be $1$, $2$ or $4$.

If an element of $G$ has order $4$, $G$ is generated by that element and therefore cyclic, and therefore abelian.

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

The identity is the only element with order 1, so all the other elements must have order $2$.

It follows directly from All Elements Order 2 then Abelian that $G$ is abelian.

The result follows.