External semidirect product

Definition with the left action convention
Suppose $$N$$ is a group and $$H$$ is a group acting on $$N$$; in other words, there is a group homomorphism $$\rho:H \to \operatorname{Aut}(N)$$, from $$H$$ to the automorphism group of $$N$$. The external semidirect product $$G$$ of $$N$$ and $$H$$, denoted $$N \rtimes H$$ is, as a set, the Cartesian product $$N \times H$$, with multiplication given by the rule:

$$\! (a,b)(a',b') = (a(\rho(b)(a')),bb')$$

Writing the action $$\rho(b)a' = b \cdot a'$$, we get:

$$(a,b)(a',b') = (a(b \cdot a'),bb')$$

The way multiplication is defined, it turns out that:


 * $$N$$ embeds as a normal subgroup of $$G$$ (via $$a \mapsto (a,e)$$) and $$H$$ embeds as a subgroup via $$b \mapsto (e,b)$$. The two subgroups are permutable complements, hence the external semidirect product is the same as an internal semidirect product once we identify $$N$$ and $$H$$ with their images in $$G$$. In particular, the image of $$N$$ is a complemented normal subgroup in $$G$$ and the image of $$H$$ is a retract of $$G$$.
 * The action of the image of $$H$$, on the image of $$N$$, via conjugation in $$G$$, is the same as the abstract action that we started with.

Case of abelian normal subgroup
In the special case where $$N$$ is an abelian group and the binary operation of $$N$$ is denoted additively, the multiplication rule for $$G$$ can be written as:

$$\! (a,b)(a',b') = (a + (b \cdot a'), bb')$$

This notation comes up in the study of the second cohomology group.

Case of trivial action
The external semidirect product becomes an external direct product when the action of $$H$$ on $$N$$ is trivial.

Related notions for groups

 * External direct product (corresponding internal notion: internal direct product, equivalence)
 * External wreath product (corresponding internal notion: internal wreath product)

Generalizations to other algebraic structures

 * External semidirect product of semigroup and group
 * External semidirect product of Lie rings