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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    $\begingroup$ What do you mean "Why do we only care about the action of conjugation"? That's the action we consider for internal direct products, but for external ones we consider any action, whether it is given by conjugation or not; that's what $\phi$ is: an arbitrary action. Conjugation only makes sense when $A$ and $B$ are subgroups of the same $G$. Moreover, in the external direct product, you have an internal semidirect product of $A\times\{1\}$ by $\{1\}\times B$, where the action by conjugation of $\{1\}\times B$ on $A\times\{1\}$ exactly corresponds to the original action $\phi$. $\endgroup$ – Arturo Magidin Jul 31 '11 at 20:09
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    $\begingroup$ Note that in the internal direct product, we want the operation to be induced by the operation of the original group $G$; that means that we want to write $(ab)(a'b')$ as a product $a''b''$, and one way to achieve this is by taking $$(ab)(a'b') = (ab)a'(b^{-1}b)b' = a(ba'b^{-1})b',$$so that is how the action by conjugation comes into the *semidirect* product. However, if $AB=BA$ as sets, even if $B$ does not normalize $A$ nor $A$ normalizes $B$, we still consider the product (it's called a product of groups) just not as a semidirect product, because neither subgroup is normal. $\endgroup$ – Arturo Magidin Jul 31 '11 at 20:11
  • $\begingroup$ @Arturo Magidin: yes the action $\phi$ is arbitrary but my question is that when we talk about "subgroups of the same group" i always see external semidirect products of the two subgroups corresponding to conjugation as it is a generic case maybe? for example consider two subgroups $A$ and $B$ of the same group $G$, $A$ normal and $A\cap B=1$ what are all split extensions of $B$ do we consider only $A\rtimes_\phi B$ with $\phi$ conjugation or we take all actions $\phi$? $\endgroup$ – palio Jul 31 '11 at 20:26
  • $\begingroup$ @Arturo Magidin: also in my definition of internal semidirect product i did not mention any action in the definition. But i said that when we consider extenal semidirect product with conjugation action then this is a group isomorphic to the internal semidirect product $AB$. $\endgroup$ – palio Jul 31 '11 at 20:30
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Aug 1 '11 at 6:21

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.


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