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I am trying to classify nonabelian groups $G$ of order 12.

I already know that they are $A_4$ (alternating group), $D_6$ (dihedral) and $Dic_{12}$ (dicyclic).

I have proved that if $n_3 \not = 1$ then $G = A_4$.

Now I am in the case where $n_3 = 1$. Then the Sylow 3-subgroup is normal in $G$, and $Aut(P_3) \simeq C_2$. And $n_2 = 3$ and $N_G(P_2) = P_2$ for each Sylow 2-subgroup. I can see that the Sylow 2-subgroups have order 4 so must be $C_4$ or $C_2 \times C_2$.

So I think that the group that arise in this case will be exactly:

  • $C_3 \rtimes C_4$
  • $C_3 \rtimes (C_2 \times C_2)$

I believe that the conditions for an internal semidirect product are:

  • We have two subgroups $N,H \le G$.
  • $N \lhd G$
  • $N \cap G$
  • $NH = G$

How can I complete this proof, showing that there are just these two semidirect product possibilities occurring?

Do I need to check all possible homomorphisms $P_2 \to Aut(C_3) \simeq C_2$ and somehow show they all lead to the same group? This seems very hard so I'd like to avoid this if possible.

Shaun
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    You only need to count homomorphisms up to $\mathrm{Aut}(P_2)$-conjugacy, as if $\phi_1$ and $\phi_2$ are homomorphisms and there exists $\sigma\in\mathrm{Aut}(P_2)$ such that $\phi_1=\sigma\phi_2$, then $\phi_1$ and $\phi_2$ determine isomorphic groups. This should be enough for you. – David A. Craven Aug 16 '20 at 21:24
  • What David said. Some of the story explained "locally" here. – Jyrki Lahtonen Aug 16 '20 at 21:49
  • I should also point out that knowing when two different extensions give you isomorphic groups, the extension problem, is really hard in general. What is even harder is when two different extensions with different normal subgroups and quotients give you the same group. – David A. Craven Aug 16 '20 at 22:07

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