Group of Order less than 6 is Abelian

Theorem
All groups with less than 6 elements are abelian.

Proof

 * 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 $$x \circ y = x \Longrightarrow y = e$$, and $$x \circ y = y \Longrightarrow x = e$$, 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 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.

No element can have order 4, for then $$G$$ would be generated by that element and therefore cyclic, and therefore abelian.

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.