Given two groups A,B. When is it possible to construct the semidirect product $A \rtimes B$.

Is there a classification for those groups?

How many semidirect products for A,b are there?

  • $\begingroup$ Where is this question coming from? Is this homework? $\endgroup$ – k.stm Apr 25 '14 at 23:19
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    $\begingroup$ A semidirect products is given by a homomorphism $\varphi : H → Aut(N)$, so the semidirect products are classified by such. $\endgroup$ – Ma Ming Apr 25 '14 at 23:19

Given two groups $A, B$ all one needs to define a semidirect product is a homomorphism $\varphi : B \rightarrow \text{Aut}(A)$. This gives the (external/outer) semidirect product $A \rtimes_{\varphi} B$.

As a set, $A \rtimes_{\varphi} B = A \times B$. The group operation is given by:

$$ (a_1,b_1)*(a_2,b_2) = (a_1 \varphi(b_1)(a_2) , b_1b_2)$$

Conversely, every semidirect product defines such a homomorphism into the automorphism group...

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As pointed out in the comment, the post, as it was, described the convention for $A\, {}_φ{\ltimes} B$, not for $A \rtimes_φ B$. So I changed that which explains the difference to @ah11950’s answer.

The semi-direct product can be constructed if you have a group homomorphism $φ \colon A → \operatorname{Aut}(B)$ by setting $A \ltimes B = A × B$ with the group structure given by $(a,b)·(a',b') = (aa',φ(a')(b)b')$.

The motivation behind this is to interpret $φ(a')$ as the conjugation by $a'$, i.e. as $b ↦ a'^{-1}ba'$. And then the equation sort of becomes $ab·a'b' = aa'·(a'^{-1}ba')b' = aa'·φ(a')(b)b'$.

Also I think, this is the convention for semi-direct products writtes as $A \, {}_φ{\ltimes} B$, i.e. the bar and the $φ$ is on the side of the “non-normal subgroup”. Not sure about this, though.

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  • $\begingroup$ You are describing the wrong convention. en.wikipedia.org/wiki/Semidirect_product $\endgroup$ – Ma Ming Apr 25 '14 at 23:29
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    $\begingroup$ Why do you think that, @MaMing ? Whether the action by automorphisms is in the first or the second coordinate is a tiny detail...or you meant something else? $\endgroup$ – DonAntonio Apr 26 '14 at 0:08
  • $\begingroup$ Well, there certainly are some suspicious notations here, indeed...but also in the other answer. $\endgroup$ – DonAntonio Apr 26 '14 at 0:09
  • $\begingroup$ @DonAntonio Your $B$ is the group to be acted; that is, the normal subgroup in the semidirect product. So the it should be written as $A\ltimes B$ or $B\rtimes A$. $\endgroup$ – Ma Ming Apr 26 '14 at 0:10
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    $\begingroup$ Yes @MaMing, I see what you mean...still, a not-so-important detail, imo. $\endgroup$ – DonAntonio Apr 26 '14 at 0:25

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