inner semidirect and outer semidirect relationship

Question

https://kconrad.math.uconn.edu/blurbs/grouptheory/group12.pdf.

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?

Answer 1

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.

Answer 2

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 $$(123)^{(12)}=(213)\neq(123).$$

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