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 $\rho :H\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$, from $H$ to the automorphism group of $N$. The external semidirect product $G$ of $N$ and $H$, denoted $N⋊H$ is, as a set, the Cartesian product $N\times H$, with multiplication given by the rule:

$(a,b)(a^{\prime },b^{\prime })=(a(\rho (b)(a^{\prime })),b{b}^{\prime })$

Writing the action $\rho (b)a^{\prime }=b\cdot a^{\prime }$, we get:

$(a,b)(a^{\prime },b^{\prime })=(a(b\cdot a^{\prime }),b{b}^{\prime })$

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.

Example

We will construct the smallest odd-order non-abelian group, the Frobenius group of order 21, as a semidirect product ${\mathbb{Z}}_7⋊{\mathbb{Z}}_3$.

We need to find a group homomorphism $\rho :{\mathbb{Z}}_3\to \mathrm{A}\mathrm{u}\mathrm{t}({\mathbb{Z}}_7)\cong {\mathbb{Z}}_6$.

Consider the map $x\mapsto (y\mapsto 2^{x}y)$

Here, $x$ is an element of the integers mod $3$, and $y$ is an element of the integers mod $7$.

This is a homomorphism, since $\rho (x{x}^{\prime })=y\mapsto 2^{x{x}^{\prime }}y=(y\mapsto 2^{x}y)\circ (y\mapsto 2^{x^{\prime }}y)=\rho (x)\rho (x^{\prime })$. Indeed, $\rho (1)=y\mapsto y$, the identity element of $\mathrm{A}\mathrm{u}\mathrm{t}({\mathbb{Z}}_7)$.

Then $G={\mathbb{Z}}_7{⋊}_{\rho }{\mathbb{Z}}_3$ is non-abelian since $(0,1)*(1,0)=(2,1)$, $(1,0)*(0,1)=(1,1)$.

Hence we have constructed a non-abelian group of order $21$.

The classification of groups of order 21 says that there is only one non-abelian group of order $21$, the Frobenius group, hence this is the Frobenius group.

Comments

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:

$(a,b)(a^{\prime },b^{\prime })=(a+(b\cdot a^{\prime }),b{b}^{\prime })$

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

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