I have been looking for a short proof of the problem
There are exactly five groups of order 12: three non-abelian and two abelian.
I know the proof of the statement, but I want to know if there is a short proof without considering all the possible relations of each element of the group with others.
EDIT: To let a brief explanation of the proof I know about this problem, what I did was consider an element $y$ of order $3$ and an element $x$ of order $4$, then consider the set generated by them: $$\{e,x,x^2,x^3,y,yx,yx^2,yx^3,y^2,y^2x,y^2x^2,y^2x^3\}$$
Then what I did was relate them with the elements of structure $y^mx^t$, that is reduced considering that some elements could not be equal considering their orders or the elements that are eliminated between them.
Well..the abelian ones are all products of appropriate cyclic groups; those aren't too tough to enumerate.
For non-abelian, the Sylow theorems tell you a lot about groups of order $p^2 q$, where $p$ and $q$ are prime: basically they're all semi-direct products, and there just aren't too many homomorphisms to use in forming the product, so it's pretty straightforward. If you don't know the Sylow theorems or semi-direct products...it's tougher.
Look especially at Dylan Moreland's answer to this question.