Group theory MOC

Semidirect product

The semidirect product 𝑁 ⋊𝐴 of groups is a generalization of the internal and external direct product of groups where only one of the operands¹ is required to be a normal subgroup of the resulting group. Semidirect products are a special case of group extension, called a ^split

1→𝑁↪𝑁⋊𝐴↠𝐴→1

since the epimorphism splits (in fact all split extensions have this form up to equivalence).

Internal semidirect product

The simpler characterization is for the internal construction. Let 𝑁 ⊴𝐺 and 𝐴 ≤𝐺 be subgroups, the first of which is normal, such that 𝑁 ∩𝐻 ={𝑒} and 𝑁𝐻 =𝐺. Then 𝐺 is the internal semidirect product 𝑁 ⋊𝐴. group

External semidirect product

For the external construction, let 𝑁 be a group and let 𝐴 be a group acting on 𝑁 by automorphisms, i.e. equipped with a homomorphism 𝜑− :𝐴 →Aut⁡(𝑁). Then the external semidirect product 𝑁 ⋊𝜑𝐴 is the set 𝑁 ×𝐴 with group multiplication given by group

(𝑛,𝑎)•(𝑚,𝑏)=(𝑛𝜑𝑎(𝑚),𝑎𝑏)

the identity is 𝑒 =(𝑒,𝑒), and the inverse is (𝑛,𝑎)−1 =(𝜑𝑎−1(𝑛−1),𝑎−1).

Proof of group

For associativity, note

((𝑛,𝑎)•(𝑚,𝑏))•(𝑙,𝑐)=(𝑛𝜑𝑎(𝑚),𝑎𝑏)•(𝑙,𝑐)=(𝑛𝜑𝑎(𝑚)𝜑𝑎𝑏(𝑙),𝑎𝑏𝑐)=(𝑛𝜑𝑎(𝑚𝜑𝑏(𝑙),𝑎𝑏𝑐)=(𝑛,𝑎)•(𝑚𝜑𝑏(𝑙),𝑏𝑐)=(𝑛,𝑎)•((𝑚,𝑏)•(𝑙,𝑐))

as required. For identity, note

(𝑛,𝑎)•(𝑒,𝑒)=(𝑛,𝑎)=(𝑛,𝑎)•(𝑒,𝑒)

as required. For inverse, note

(𝑛,𝑎)•(𝜑𝑎−1(𝑛−1),𝑎−1)=(𝑛𝜑𝑎(𝜑𝑎−1(𝑛−1)),𝑎𝑎−1)=(𝑛𝜑𝑎𝑎−1(𝑛−1),𝑒)=(𝑛𝑛−1,𝑒)=(𝑒,𝑒)(𝜑𝑎−1(𝑛−1),𝑎−1)•(𝑛,𝑎)=(𝜑𝑎−1(𝑛−1)𝜑𝑎−1(𝑛),𝑎−1𝑎)=(𝜑𝑎−1(𝑛−1𝑛),𝑒)=(𝜑𝑎−1(𝑒),𝑒)=(𝑒,𝑒)

as required.

Right action convention

If we instead have right actions, we define the product by 𝑁 ⋊𝜑𝐻

(𝑛1,ℎ1)(𝑛2,ℎ2)=(𝑛1𝑛(ℎ−11)𝜑2,ℎ1ℎ2)

for 𝑛𝑖 ∈𝑁 and ℎ𝑖 ∈𝐻 with

(𝑛,ℎ)−1=((𝑛−1)(ℎ)𝜑,ℎ−1)

for 𝑛 ∈𝑁, ℎ ∈𝐻.

Relationship between internal and external semidirect product

If 𝐺 is the internal semidirect product 𝑁 ⋊𝐴, then 𝐺 is isomorphic to the external semidirect product 𝑁 ⋊𝜑𝐴, group where 𝜑 denotes the conjugation action (which leave 𝑁 invariant by normality).

Likewise, if 𝐺 is the external semidirect product 𝑁 ⋊𝜑𝐴, then

• the subset 𝑁𝐺 =𝑁 ×{𝑒𝐴} ⊆𝐺 is a normal subgroup isomorphic to 𝑁
• the subset 𝐴𝐺 ={𝑒𝑁} ×𝐴 ⊆𝐺 is a subgroup isomorphic to 𝐴
• 𝐺 is the internal semidirect product 𝑁𝐺 ⋊𝐴𝐺
• Conjugation of an element of 𝑁𝐺 by an element of 𝐴𝐺 is the group action 𝜑.

Hence if the action 𝜑 is trivial, then the semidirect product coïncides with the direct product of groups.

Proof

Let 𝑁 ⊴𝐺 be a normal subgroup and 𝐴 ⊆𝐺 be subgroups such that 𝐺 =𝑁𝐴 and 𝑁 ∩𝐴 ={𝑒}. Since 𝑁 is normal it is invariant under Inn(𝐺), so conjugation by elements of 𝐴 is a group action on 𝑁, which we denote 𝜑. Let ˜𝐺 =𝑁 ⋊𝜑𝐴, and Ψ :˜𝐺 →𝐺 :(𝑛,𝑎) ↦𝑛𝑎. Then for any 𝑛,𝑚 ∈𝑁 and 𝑎,𝑏 ∈𝐴

Φ((𝑛,𝑎)•(𝑚,𝑏))=Φ(𝑛𝜑𝑎𝑚,𝑎𝑏)=Φ(𝑛𝑎𝑚𝑎−1,𝑎𝑏)=𝑛𝑎𝑚𝑎−1𝑎𝑏=𝑛𝑎𝑚𝑏=Φ(𝑛,𝑎)Φ(𝑚,𝑏)

hence Φ is a homomorphism, in particular it is an epimorphism 𝑁𝐴 =𝐺. Now let 𝑛 ∈𝑁 and 𝑎 ∈𝐴 such that Φ(𝑛,𝑎) =𝑛𝑎 =𝑒. It follows that 𝑛 =𝑎−1, so both 𝑁 and 𝐴 must contain 𝑛, which can only be true if 𝑛 =𝑎 =𝑒. Therefore Φ is a monomorphism and thus an isomorphism.

Now let 𝑁,𝐴 be arbitrary groups, 𝜑− :𝐴 →Aut⁡(𝑁) be a (left) group action of 𝐴 on 𝑁, and 𝐺 =𝑁 ⋊𝜑𝐴. Furthermore let 𝑁𝐺 =𝑁 ×{𝑒} and 𝐴𝐺 ={𝑒} ×𝐴 as sets. Since (𝑛,𝑒) •(𝑚,𝑒) =(𝑛𝑚,𝑒) for any 𝑛,𝑚 ∈𝑁, and likewise (𝑒,𝑎) •(𝑒,𝑏) =(𝑒,𝑎𝑏), it is clear that 𝑁𝐺 and 𝐴𝐺 are subgroups of 𝐺 isomorphic to 𝑁 and 𝐴 respectively.

Note that

𝑁𝐺𝐴𝐺=(𝑁,𝑒)•(𝑒,𝐴)=(𝜑𝐴𝑁,𝐴)=(𝑁,𝐴)

so 𝐺 =𝑁𝐺𝐴𝐺.

Let 𝑛 ∈𝑁 and 𝑎 ∈𝐴. Then

(𝑒,𝑎)•(𝑛,𝑒)•(𝑒,𝑎)−1=(𝑒,𝑎)•(𝑛,𝑒)•(𝑒,𝑎−1)=(𝜑𝑎(𝑛),𝑎)•(𝑒,𝑎−1)=(𝜑𝑎(𝑛),𝑒)

as claimed above. This also shows that 𝑁𝐺 ⊴𝐺, since conjugating by any element is equivalent to conjugating by an element of 𝑁, and then conjugating by an element of 𝐴. Clearly 𝑁𝐺 ∩𝐴𝐺 ={(𝑒,𝑒)}, so 𝐺 =𝑁𝐺 ⋉𝐴𝐺 internally.

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Footnotes

1. that to which the triangle points, so 𝑁 is normal in 𝑁 ⋊𝐴 and 𝐴 ⋉𝑁. ↩