Prove every group of order less or equal to five is abelian Is it possible to prove that every group of order less or equal to five is abelian?   
We know that groups of prime order are cyclic and therefore commutative. As the number $4$ is the only composite number $\le5$, it basically remains to show this for groups of order four.

Another way to prove it is that a finite ordered non-abelian group must have order 6 or greater which can be done like this
if $|G|=1$ then it's trivial, So assume $|G|>1$ then there exist $a\in G$, but all groups have inverses so $a^{-1}\in G$, which means $|G|\geq 3$, but this case would be abelian so there must be another element $b\in G$ and equally so it's inverse, so we have that $|G|\geq 5$, But we also have that $ab\in G$, which would give one of the following $ab=a$, $ab=b$, $ab=e$, $ab=a^{-1}$ or $ab=b^{-1}$, the first ones can be eliminated as they imply that either $a$ or $b$ are identity or inverse, the latter two can be eliminated for the reason that they would make it abelian as we get $aba=a^2b=e$, with left and right a multiplication on $ab=a^{-1}$. This gives us that $ab=ba$ for the elements which clearly doesn't work as it is a non-abelian group and the remainder would be the same (with the inverses etc), so there must exist some other element then. So $|G|\geq 6$
