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Currently I am going through Representation theory of semi-direct products by Reyes. At the begining of the article, the author defines the semidirect product as follows.

$G = H \cdot B$ is a semidirect product where $H$ is a normal subgroup of $G$ and $B$ is isomorphic with a group of automorphisms of $H$.

This doesn't match with the definition given here which is:

Semidirect products.

We say that $G$ is (isomorphic to) a semidirect product of $M$ by $N$ if and only if there exist subgroups $H$ and $K$ of $G$ such that:

  • $H\cong M$ and $K\cong N$;
  • $H\triangleleft G$;
  • $H\cap K=\{e\}$;
  • $G=HK$.

Because, in the second definition $H$ is not required to be isomophic with a group of automorphism of $K$. So, can I say that the first definition is not a general definition of semidirect product?

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See "outer" and "inner" semi-direct products in https://en.wikipedia.org/wiki/Semidirect_product

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  • $\begingroup$ so it looks, Reyes was talking about the outer semidirect product, right? $\endgroup$ – Omar Shehab Oct 23 '16 at 22:40
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    $\begingroup$ Yes, and the two constructions are closely related. $\endgroup$ – David Towers Oct 24 '16 at 7:19
  • $\begingroup$ could you please explain how they are related? $\endgroup$ – Omar Shehab Oct 24 '16 at 18:15
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    $\begingroup$ A group $G$ is an "inner" semidirect product of $H$ and $K$ if and only if it is isomorphic to an "outer" semidirect product. "Outer" means giving an actual construction of the semidirect product; "inner" means recognizing it in the wild. $\endgroup$ – Ravi Fernando Oct 25 '16 at 3:39
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    $\begingroup$ Also a slight comment on Reyes's line: when talking about (outer) semidirect products, we need a homomorphism from $K = B$ to $Aut(H)$, but this homomorphism does not need to be injective; i.e. it does not need to identify $K$ as a subgroup of $Aut(H)$. For example, a direct product $H \times K$ is the case where $K \to Aut(H)$ is the trivial homomorphism. This is definitely a special case of semidirect products, but it seems like a stretch to call $K$ "isomorphic with a group of automorphisms of $H$" when they all act as the identity. $\endgroup$ – Ravi Fernando Oct 25 '16 at 3:43

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