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take $A$ and $B$ two subgroups of THE SAME group $G$. In this post i'm not interested in the general case of $A$ and $B$ being given groups not necessarely subgroups of the same group.

if $A\cap B =\{1\}$ and $A$ is normalized by $B$ then the set $AB=\{ab\,|a\in A, b\in B\} $ is a subgroup of $G$ that we call the internal semidirect product of $A$ by $B$.

Now for each action $\phi:B\rightarrow Aut(A)$ we can endow the cartesian product $A\times B$ with the group structure $(a_1,b_1)*(a_2,b_2)=(a_1\phi_{b_1}(a_2),b_1b_2)$ and we get a group called the external semidirect product of $B$ acting on $A$ corresponding to $\Phi$. Why we only care about the action of conjugation $\phi_b(a)=bab^{-1}$? I know only one reason: when acting by conjugation, the external semidirect product $A\times B$ is ISOMORPHIC to the internal semidirect product $AB$. But what about other actions? are there situations where external semidirect products of subgroups $A$ and $B$ of the same group $G$ with action other than conjugation are considered?

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    There are *two different question*: What are all split extension of $A$ by $B$? That's one question. If $A$ and $B$ are *subgroups of the same group $G$*, with $A$ normal, what are the actions of $B$ on $A$? That's a **different** question; the only action that makes sense *in that context* is conjugation. "What are all split extension of $B$ do we consider only..." does not parse.2011-08-01

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In the definition of external semidirect product, you can take $\phi$ to be an arbitrary action and you still get a group. Then, $\phi$ turns out to be just conjugation inside the external semidirect product (note that "conjugation" action doesn't make any sense unless you say conjugation inside which group). So, really, all actions are conjugation actions, since they are conjugation inside the external semidirect product.