# External semidirect product

## Definition

### 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.