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Give all groups of cardinality 12.

What I did:

let G be such a group.

$12 = 2^2.3$ so G has a 2-Sylows $S_2$ and a 3-Sylows $S_3$.

I proved that either $S_2$ or $S_3$ is normal and $S_3\cap S_2=\{e\}$. Therefore: $G\cong S_2\rtimes S_3$ or $G\cong S_3\rtimes S_2$.

We have $S_2\cong \Bbb Z/4\Bbb Z$ or $S_2 \cong (\Bbb Z/2\Bbb Z)^2$ and $Aut(S_3) = Aut(\Bbb Z/4\Bbb Z)=\Bbb Z/2\Bbb Z$.

  1. Using Chinese lemma, I found the following direct products, $\Bbb Z/4\Bbb Z \times \Bbb Z/3\Bbb Z\cong \Bbb Z/12\Bbb Z,\ \Bbb Z/2\Bbb Z \times \Bbb Z/6\Bbb Z,\ \Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z \times \Bbb Z/3\Bbb Z$.
  2. I don't know how to compute $Aut((\Bbb Z/2\Bbb Z)^2)$ in order to find non trivial morphisms $S_3 \to Aut((\Bbb Z/2\Bbb Z)^2)$ and I have trouble finding non trivial morphisms $S3 \to Aut(S_2)$ and $S_2 \to Aut(S_3)$ in order to find the semidirect products.

Thank you for your help and comments

Conjecture
  • 3,088
  • One note: $\Bbb Z/2 \Bbb Z \times \Bbb Z \ 6 \Bbb Z \cong \Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z \times \Bbb Z / 3 \Bbb 3 \Bbb Z$. – Robert Shore Mar 12 '19 at 15:31
  • Possible duplicate of Finding and classifying all groups of order 12. – Dietrich Burde Mar 12 '19 at 15:31
  • In the supposed duplicate question there is no mention of semi direct products and only a case is discussed. – Conjecture Mar 12 '19 at 15:55
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    Indeed the Accepted Answer to the proposed duplicate, despite its length and having upvotes, seems primarily a plea for help in better understanding material that was "gone over in class." The reason is that the body of the Question was copied and some clarifying remarks were interpolated, thereby getting the length. I'm voting to reopen, though perhaps a better duplicate can be found. – hardmath Mar 12 '19 at 16:39
  • I have now the full answer to my question, There are 5 groups up to an isomorphism, and I would like to share it with everyone but this question is mistakingly marked as duplicate. @DietrichBurde: Please reopen this question so I can answer this question once and for all, it may be helpful for other people like me in the future. Otherwise, please provide a better duplicate. – Conjecture Mar 12 '19 at 18:58
  • There are indeed several other duplicates and references to books, which give better proofs, because they have chapters before explaining more details with more examples. I think everything has been said about groups of order $12$ for this purpose. So no need to rewrite this again. – Dietrich Burde Mar 12 '19 at 19:03

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