In the text above, the author explains how to find all groups of order 12.

He does so by showing that a group $G$ (of order $12$) is isomorphic to the semi-direct product of the $2$-sylow ($P_2$) and $3$-sylow ($P_3$) subgroups.

Then, he splits to cases and goes through all the possible combinations of semidirect products of $P_2$ ($\Bbb Z_4$ or $\Bbb Z_2 \times \Bbb Z_2$) with $P_3$ ($\Bbb Z_3)$.

Now my problem arises: he also looks for all possible actions $P_3 \to {\rm Aut}(P_2)$ (or $P_2 \to {\rm Aut}(P_3)$).

But why is this necessary?

He proved that $P_2 \cap P_3$ is trivial, $|P_2P_3| = |G|$ and that $P_2$ is normal or $P_3$ is normal. This means that $G$ is isomorphic to the inner semi-direct product, so why there could be more nonisomorphic semidirect products?

An example of what I'm trying to ask:
Let's say we're in the case that $P_2$ is normal and isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, he found that $G$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \rtimes \Bbb Z_3$ or $G$ isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3$.
But, we already know that $G$ is isomorphic to the inner semidirect product, so how is it possible that there are two options but not one?


For any action $\varphi:G\to {\rm Aut}(H)$ we obtain a semidirect product by defining multiplication on elements of $G\times H$ by $$(g,h)(g',h'):=(gg',\,\varphi_{g'}(h)h')$$ where conjugation $g^{-1}hg$ becomes $\varphi_g(h)$ because $$(g^{-1},1)(1,h)(g,1)=(1,\varphi_g(h))\,.$$ For distinct actions, we can indeed obtain nonisomorphic semidirect products of the same two groups. For instance, if $\varphi$ is the constant map ${\rm id}_H$, then we obtain the ordinary direct product as a special case.

  • $\begingroup$ Thank you, but I'm not sure what you tried to explain, can you please elaborate? Also, can you please look at the "example" in the end? Where is my mistake there? $\endgroup$ – yong Jan 9 at 23:01
  • $\begingroup$ The inner semiproduct structure alone doesn't determine the group up to isomorphism, only together with the given (inner) automorphisms $h\mapsto g^{-1}hg$. $\endgroup$ – Berci Jan 10 at 0:22
  • $\begingroup$ that's what I meant - in the example I gave, G should be isomorphic only to the inner semidirect product with the inner automorphisms - but in reality G might be isomorphic to 2 groups (depending on the actions). $\endgroup$ – yong Jan 10 at 16:05

Consider, for example, the cyclic group $\langle a\rangle$ of order six. Then

$$\langle a\rangle =\langle a^2\rangle\rtimes \langle a^3\rangle=\{1,a^2,a^4\}\times \{1,a^3\}.$$

But the symmetric group

$$S_3=\langle (123)\rangle\rtimes\langle (12)\rangle=\{{\rm id},(123),(132)\}\rtimes\{{\rm id}, (12)\}.$$

So nonisomorphic groups can be written as semidirect product, whose corresponding factors are isomorphic!

The distinction, as Berci points out, is in conjugation. We have $$(a^2)^{a^3}=a^2$$ but $$(123)^{(12)}=(213)\neq(123).$$

I found this example in Roman's "Fundamentals of Group Theory: An Advanced Approach" early on in the section on semidirect products.


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