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
Best Answer
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.