What is the smallest order of a group to have non abelian proper subgroup? What is the smallest order of a group to have non abelian proper subgroup?
Is there any useful theorem or trick to answer this question.
I couldn't find any useful theorem to solve this problem in my book. This question is from my recent exam.
 A: Let $G$ be the answer group.  Clearly, $G$ must be non-abelian.  The "first" one is $S_3$, but this won't work since every proper non-trivial subgroup has order 2 or 3.  Next up are the non-abelian groups of order 8, but here a proper non-trivial subgroup has order 2 or 4, but these are all abelian (cyclic or Klein-4).  With order 10 you only have prime divisors 2 and 5, so it's much the same story.  The next non-abelian groups are those of order 12, and sure enough, $G=S_3 \times \mathbb{Z}_2$ works. 
Edit:  I skipped "obvious" orders, like 7, 9, and 11 in which there are no non-abelian types. 
A: You can see that the smallest non abelian group has order 6. So if you want a group that has a non abelian proper subgroup, its order has to be at least 12. Consider the direct product of that non abelian group of order 6 and  $\mathbb{Z}/2\mathbb{Z}$.
A: Do you know what the smallest nonabelian group is?  A good candidate would be the smallest group that it is a subgroup of.  You know the order of a subgroup divides the order of a group.
