Finding and classifying all groups of order 12

Question

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

Answer 1

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.

Answer 2

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 $A_4$.

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