Semidirect product

Definition a remarks

We start with a proposition:
Proposition
Let $G$ be a group, $N$ a normal subgroup and $H$ a subgroup. The following are equivalent:

There exists a group homomorphism $ϕ : G \to H$ such that $ϕ |_{H} = i d$ and $Ker (ϕ) = N .$
It is satisfied that $G = N H$ and $H \cap N = {e} .$

When any of those requirements is satisfied we write

G = N ⋊ H

$◼$
Definition
When we have the context given by condition 1 or 2 we say that $G$ is the semidirect product of $N$ and $H$ and we write $G = N ⋊ H$ .
$◼$

Remarks

The "converse" is not true. Given a normal subgroup $N$ of $G$ , $G$ is not necessarily the semidirect product of $N$ and $G / N$ . For example: $Z / 4 Z$ which contains $Z / 2 Z$ . I have to think it.
Another characterization. An exact sequence $0 \to N \to G \to H \to 0$ to be isomorphic to one of the form $0 \to N \to N ⋊ H \to H \to 0$ is if $G \to H$ has a section.
Here $N H$ must not be confused with $N \times H$ , the direct product (see product (cat th)). Instead, is a product of group subsets,

N H = {n h : n \in N, h \in H} .

Under condition 2 every $g \in G$ can be written in an unique way as a product $g = n h$ , $n \in N, h \in H$ . If $n h = n^{'} h^{'}$ then $(n^{'})^{- 1} n = h^{'} h^{- 1} \in N \cap H$ and therefore $(n^{'})^{- 1} n = h^{'} h^{- 1} = e .$
Condition 1 is like saying that the main group $G$ can be seen like a bundle, since we have a idempotent map $ϕ : G \to H$ . The standard fibre is $N$ .

Image from Mathematics for Physics, an ilustrated handbook, page 40.

Proof
Pasted image 20211230180557.png
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The product in $G$

The product in $G$ takes a special form: given $g = (n, h)$ and $g^{'} = (n^{'}, h^{'})$ then $g g^{'} = (\bar{n}, \bar{h})$ and

\bar{n} \bar{h} = n h n^{'} h^{'} = n h n^{'} h^{- 1} h h^{'}

therefore

\bar{h} = h h^{'}

\bar{n} = n h n^{'} h^{- 1}

We can think in $N$ as translations and $H$ as rotations.

Key example

Suppose the group $E (2)$ of isometries of the euclidean plane, or rigid motions. It is known that any element of $E (2)$ can be seen as the composition of a translation (an element of $R^{2}$ ) and a linear isometry (an element of $O (2)$ , that is, a rotation with centre at the origin or a reflection with axis through the origin). This decomposition is unique.

Moreover, translations are a normal subgroup of $E (2)$ : you can see it intuitively "moving things on a table" or by computation:

v \mapsto R (R^{- 1} v + t) = R R^{- 1} v + R t = v + R t

Therefore

E (2) = R^{2} ⋊ O (2) .

In general, if we now consider any subgroup $H$ of $G L (n)$ we can consider the subgroup $G$ of $G L (n + 1)$

G = {(\begin{matrix} B & v \\ 0 & 1 \end{matrix}) : B \in H, v \in R^{n}}

It is easy to show that $G$ is isomorphic to $R^{n} ⋊ H$ (construct the map $ϕ$ of the definition).

Conversely, given any group $G = N ⋊ H$ , with $N \approx R^{n}$ and $H \approx \tilde{H} \leq G L (n)$ , it can be shown that $G$ is isomorphic to a subgroup of $G L (n + 1)$ of this form. Consider

\bar{N} = {(\begin{matrix} I & v \\ 0 & 1 \end{matrix}) : v \in R^{n}}

and

\bar{H} = {(\begin{matrix} B & 0 \\ 0 & 1 \end{matrix}) : B \in \tilde{H}}

Now it is easy to see that $G$ is isomorphic to a subgroup of $G L (n + 1)$ with the form above by means of the map:

φ : G = N ⋊ H \to \bar{N} \times \bar{H} \to G L (n + 1)

(n, h) \mapsto ((\begin{matrix} I & v \\ 0 & 1 \end{matrix}), (\begin{matrix} B & 0 \\ 0 & 1 \end{matrix})) \mapsto (\begin{matrix} B & v \\ 0 & 1 \end{matrix})

This map is a homomorphism. The product in $G$ takes the form

(n, h) (n^{'}, h^{'}) = (n h n^{'} h^{- 1}, h h^{'})

and with a standard computation can be checked that

φ ((n, h)) \cdot φ ((n^{'}, h^{'})) = φ ((n h n^{'} h^{- 1}, h h^{'}))

Moreover, the map is also injective, as it is easy to check.

Another approach

Within this approach we can create a group being the semidirect product of two given groups.

Definition. Let $H$ and $N$ be groups, and let $ϕ : H \to Aut (N)$ be a group homomorphism such that $ϕ (h)$ defines a group action of $h$ on $N$ . The semidirect product $N ⋊_{ϕ} H$ is the Cartesian product $N \times H$ endowed with the multiplication operation

\cdot : (n, h) \cdot (n^{'}, h^{'}) = (n ϕ (h) (n^{'}), h h^{'}) \forall n, n^{'} \in N, h, h^{'} \in H,

where $ϕ (h) (n^{'})$ denotes the action of $h$ on $n^{'}$ . $◼$

It can be shown that it is indeed a group.

About the Lie algebras

I think it can be proven the following proposition
Proposition
Let $G$ be a Lie group that is the semidirect product of a normal subgroup $N$ and a subgroup $H$ , i.e., $G = N ⋊ H$ . Then the Lie algebra $g$ of $G$ has a canonical decomposition as an $A d (H)$ -module given by

g = n \oplus h,

where $n$ and $h$ are the Lie algebras of the subgroups $N$ and $H$ , respectively. $◼$

Proof
The idea is:

Show that $g = n \oplus h$ as vector spaces
$g$ is an $A d (H)$ -module, and of course $A d (H) (h) \subset h$ . But also, since $N$ is normal, we have that $A d (H) (n) \subset n$
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