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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.

iam_agf
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  • Yes, corrected. – iam_agf Dec 17 '16 at 01:31
  • Well, how about you letting us know the proof of the statement you know; we can't give you a "short(er) proof" without knowing that. – amWhy Dec 17 '16 at 01:38
  • @amWhy I let the way of my proof, but considering that is very large, I only added the parts that are useful to understand it. – iam_agf Dec 17 '16 at 01:45
  • How you prove that they're the only ones? For the abelian case is very easy. But for the non abelian, it isn't. In fact you're missing $D_{12}$ (or $D_6$ depending of the author) – iam_agf Dec 17 '16 at 01:47
  • How do you know there is an element $x$ of order 4 in the group? (You don't, since sometimes there isn't.) –  Dec 17 '16 at 02:45
  • Actually, two abelian and three nonabelian. – anon Dec 17 '16 at 02:49

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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.

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John Hughes
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  • Damn, you're right. The semidirect product was what I was looking for the non-abelian case. – iam_agf Dec 17 '16 at 01:33
  • Yes, with this is very clear considering the semidirect products. Considering that I finished in less than an hour, yes, is very short. – iam_agf Dec 17 '16 at 01:49