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.

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

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.


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

  • $\begingroup$ Yeah, this is the fastest way to get there. $\endgroup$ – Randall Nov 1 '17 at 2:41
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    $\begingroup$ The next question someone wants to ask, could be: Are there other (up to isomorphism) groups, also of order 12, with this property? $\endgroup$ – Jeppe Stig Nielsen Nov 1 '17 at 11:49

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.


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