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I was working on classifying all the groups of order 12. I dug around at some of the previous questions here and while they address the idea, none of them were entirely satisfactory:

Classifying groups of order 12. (doesn't explain how classification is derived)

Group of order 12 (doesn't address how to generate classification)

Nonisomorphic groups of order 12. (shows they aren't isomorphic, but doesn't actually show the derivation for the 4 groups listed)

So I wanted to ask, the specific question not yet presented, of how to derive that there must be 5 non isomorphic groups of order 12, and which groups those are.

Work So Far:

To begin with we have $|G|=12= 2^2 \times 3$ for any group of order 12. Let $n_3$ be the number of sylow-3 subgroups and let $n_2$ be the number of sylow 2 subgroups. We have by the third sylow theorem that

$$n_3 | 4, n_3 \equiv 1 \mod 3$$ $$n_2 | 3, n_2 \equiv 1 \mod 2$$

So the groups can have either 1 or 4 sylow 3 subgroups of order 3, and either 1 or 3 sylow 2 subgroups of order 4.

Furthermore coprime sylow-p groups only share the identity element in common, so we can rule out 4 sylow-3 groups and 3 sylow-2 groups, as the presence of either rules out the existence of a single copy of the other.

Now we have established our groups must have a SINGLE sylow 2 subgroup, and a SINGLE sylow 3 subgroup, and that means that each is a normal subgroup of the entire group.

The sylow-2 subgroup can be either $\Bbb{Z}_2 \times \Bbb{Z}_2$ or $\Bbb{Z}_4$. And the sylow-3 subgroup has a single contender $\Bbb{Z}_3$.

Naturally then we can list out two groups

$$\Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_3$$ $$\Bbb{Z}_4 \times \Bbb{Z}_3$$

But now the question remains, how to discover any remaining groups, and show that the remaining set covers all possible groups.a

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Sidharth Ghoshal
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    This paper discusses this. – Cameron Buie Dec 20 '15 at 22:12
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    Certainly it is helpful to check MSE for existing answers. But for a question like this you should read a paper; and I find the above link excellent. – Dietrich Burde Dec 20 '15 at 22:15
  • This is a relatively small case, so I'm sure it is covered in many a textbook. I recall first reading about it from Baumslag & Chandler's book (Schaum Outline series). – Jyrki Lahtonen Dec 20 '15 at 23:38
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    You seem to be saying that a group of order $12$ must have a unique Sylow $2$-subgroup and a unique Sylow $3$-subgroup, but the dihedral group of order $12$ has $3$ Sylow $2$-subgroups and the alternating group of degree $4$ has $4$ Sylow $3$-subgroups. – Zoe H Dec 21 '15 at 00:41
  • My answer here gives a brisk but almost complete answer – Rylee Lyman Jul 07 '19 at 12:49
  • You say "the presence of either rules out the existence of a single copy of the other." As if having 4 Sylow-3 subgroups somehow eats up so many elements that there is no room for even one Sylow-2 subgroup (of order 4). But that just isn't the case. 4 Sylow-3 subgroups accounts for $1+4(2)=9$ elements. That leaves 3 elements, which together with the identity could make a Sylow-2 subgroup. – 2'5 9'2 Mar 29 '22 at 18:07
  • The link by @CameronBuie seems to have broken. This should work. (I think it's the same paper.) – Robin Dec 11 '23 at 10:47

2 Answers2

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For the non-abelian ones: By counting, one or the other Sylow subgroups is normal.

Now $P_3=\Bbb Z_3, P_2=\Bbb Z_4$ or $P_2=\Bbb Z_2×\Bbb Z_2$.

By the first observation, it easily follows that $G$ of order $12$ is a semi-direct product.

Either $G\cong \Bbb Z_3\rtimes \Bbb Z_4,$ or $\,G\cong \Bbb Z_3\rtimes (\Bbb Z_2×\Bbb Z_2)$, or $(\Bbb Z_2×\Bbb Z_2)\rtimes \Bbb Z_3$.

These groups are the dicyclic group of order $12$, $D_6$ and $A_4$.

The abelian groups are easy with the structure theorem: $$\Bbb Z_{12},\,\Bbb Z_6×\Bbb Z_2$$.

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A simple solution can be found in the book "Groups and symmetries by M.A Armstrong". It is given in the chapter sylow's theorem. you can download the book from-- http://bookzz.org/book/714875/11dbeb.

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