Once we have two groups, let's say $H$ and $G$, it's fairly simple to find all split extensions of $G$ by $H$ by looking at the automorphism group $\text{Aut}(H)$ and homomorphisms $\tau: G \rightarrow \text{Aut}(H)$. Then, the extensions are described by the semidirect products $H \rtimes G$. In a similar way, once we have a group $A$, we can find a normal subgroup on $A$ and write $A$ as a semidirect product.

My question is, on the first case, unless the homomorphism $\tau$ is fairly trivial, we end up with a group $H \rtimes G$ that we know little about. The operation defined by the external semidirect product is awkward to handle, which leaves me pretty lost when trying to define who exactly is $H \rtimes G$.

The last one I tried was looking at the following split extension: $1 \rightarrow C_3 \rightarrow G \rightarrow C_2 \times C_2 \rightarrow 1$. So $G$ is a extension of the Klein four-group by $C_3$, the cyclic group of three elements. There are a few possible homomorphisms $\tau: C_2 \times C_2 \rightarrow \text{Aut}(C_3)$, so I decided to start with a simple one, a projection: $\tau(a, b) = a$ (I used this notation because $\text{Aut}(C_3) \cong C_2$).

And that was pretty much all I could do. Sure $G$ is isomorphic to one of the five groups of order $12$, but which one? Would I need to go through all groups of order $12$, searching for one with a normal subgroup isomorphic to $C_3$, whose quotient group is isomorphic to $C_2 \times C_2$? There must be more I can extract from the external semidirect product, otherwise it would make it pretty pointless for this objective. What am I missing here?

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    $\begingroup$ It is not too hard to see that the semidirect product that you describe is isomorphic to $(C_3 \rtimes C_2) \times C_2 \cong D_6 \times C_2$, where $D_6$ is the dihedral group of order $6$. (Note also that $D_6 \times C_2 \cong D_{12}$.) The only other possibility with these groups is when $\tau$ is the trivial map, in which case you get the direct product.. But I don't have a general answer to your question! $\endgroup$ – Derek Holt Dec 23 '15 at 19:41

In your specific case, note that the action of $C_2^2$ on $C_3$ does not depend on the second coordinate, so this is actually $(C_3\rtimes C_2)\times C_2$, i.e. $S_3\times C_2$.

This is the only nontrivial split extension of $C_3$ by $C_2^2$: to define a homomorphism of $C_2^2\to \operatorname{Aut}(C_3)$, you need to define the image of each generator, and it is easy to check that the other two nontrivial homomorphisms are conjugate (by an automorphism of $C_2^2$).

If you are looking for a "general procedure", I doubt there is anything of the sort. Generally speaking, these sorts of problems in group theory are very hard, including the ones you have dismissed as simple. I don't think there is any easy way to find all maps $G\to \operatorname{Aut}(H)$ (even if we have a good description of this $\operatorname{Aut}(H)$), nor is it easy to find all normal subgroups of a given (arbitrary) group. These statements about hardness of respective problems can be made precise in the language of computability/computational complexity theory.

That said, if you are interested in finding out the result, instead of having any sort of deep understanding it, then computer algebra systems are your friend: they can compute all these things for (a large class of) finite, or even finitely presentable groups (and probably some others), and find a human-readable description of a given group. I've used GAP in the past, and I think you should have no trouble using it, as long as you are at all comfortable with programming.

But the hardness is still there: when the sizes of the groups become large (say, larger than 20 elements), expect the computation to require significant resources (in terms of computation time and memory).

  • $\begingroup$ Your answer, google and the comment by @DerekHolt led me to the isomorphism property I didn't knew about (actions that can be transformed in one another by a automorphism of the acting group lead to the same semidirect product, this is not on my books). My only open question is, we still need knowledge about the bigger groups to understand the semidirect products right? Like the step when both of you said $C_3 \rtimes C_2 \cong S_3$. This is known beforehand, by studing $S_3$, and not derived by some property of $C_3 \rtimes C_2$, right? So, I would indeed need to know the groups of order 12. $\endgroup$ – Henrique Augusto Souza Dec 23 '15 at 20:02
  • $\begingroup$ for your edit, I've been thinking on giving GAP a chance, but I would be more interested in it for checking my answers then doing calculations by itself. My real objective is to understand how the semidirect product is a useful tool for describing groups and their properties. $\endgroup$ – Henrique Augusto Souza Dec 23 '15 at 20:10
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    $\begingroup$ As for your second comment, the main strength of the semidirect product lies (I think) in the fact that if we know how to work (perform calculations) with elements of $G,H$, then chances are, we can do the same with the semidirect product (well, there is a caveat or two, but mostly...). In case the basic group is abelian, this can be improved to non-split extensions by adding a cocycle into the mix: there is a complete description of all extensions, split or not (of a given abelian group by a fixed group) in terms of the group action and the cocycle (see e.g. Rotman's Advanced Modern Algebra). $\endgroup$ – tomasz Dec 23 '15 at 22:38
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    $\begingroup$ @HenriqueAugustoSouza: As for your first comment, the part about automorphism is not all that complicated: you don't even need an automorphism: if you switch one of the group in the semidirect products with an isomorphic one (and twist everything via the isomorphism), of course you will get an isomorphic group -- after all, semidirect product is a purely group-theoretical construction. $\endgroup$ – tomasz Dec 23 '15 at 22:45
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    $\begingroup$ In the second part, of course we need to know something about $S_3$. You can't identify a group as isomorphic with another group if you do not know anything about that other group, so I don't really see what you mean by that. Maybe it'd be better to say that $C_3\rtimes C_2\cong D_6$ (because this is a natural description of $D_6$, not $S_3$ -- we have in general $C_n\rtimes C_2\cong D_{2n}$, by definition you might say, but the isomorphism $D_6\cong S_3$ is exceptional). $\endgroup$ – tomasz Dec 23 '15 at 22:46

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