# 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.

• I believe the answer is 12. – Randall Nov 1 '17 at 2:32
• No trick: just knowing the structure of small groups up to iso. – Randall Nov 1 '17 at 2:32
• Would you like to explain? – Bora Nov 1 '17 at 2:33

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 $\mathbb{Z}/2\mathbb{Z}$.
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.