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 extension

1 \to 𝑁 ↪ 𝑁 ⋊ 𝐴 ↠ 𝐴 \to 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 $𝐴 \leq 𝐺$ be subgroups, the first of which is normal, such that $𝑁 \cap 𝐻 = {𝑒}$ and $𝑁 𝐻 = 𝐺$ . Then $𝐺$ is the internal semidirect product $𝑁 ⋊ 𝐴$ . #m/def/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 $𝜑 - : 𝐴 \to A u t (𝑁)$ . Then the external semidirect product $𝑁 ⋊ 𝜑 𝐴$ is the set $𝑁 \times 𝐴$ with group multiplication given by #m/def/group

(𝑛, 𝑎) • (𝑚, 𝑏) = (𝑛 𝜑 𝑎 (𝑚), 𝑎 𝑏)

the identity is $𝑒 = (𝑒, 𝑒)$ , and the inverse is $(𝑛, 𝑎) - 1 = (𝜑 𝑎 - 1 (𝑛 - 1), 𝑎 - 1)$ .

Proof of group

For associativity, note

((𝑛, 𝑎) • (𝑚, 𝑏)) • (𝑙, 𝑐) = (𝑛 𝜑 𝑎 (𝑚), 𝑎 𝑏) • (𝑙, 𝑐) = (𝑛 𝜑 𝑎 (𝑚) 𝜑 𝑎 𝑏 (𝑙), 𝑎 𝑏 𝑐) = (𝑛 𝜑 𝑎 (𝑚 𝜑 𝑏 (𝑙), 𝑎 𝑏 𝑐) = (𝑛, 𝑎) • (𝑚 𝜑 𝑏 (𝑙), 𝑏 𝑐) = (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) • (𝜑 𝑎 - 1 (𝑛 - 1), 𝑎 - 1) = (𝑛 𝜑 𝑎 (𝜑 𝑎 - 1 (𝑛 - 1)), 𝑎 𝑎 - 1) = (𝑛 𝜑 𝑎 𝑎 - 1 (𝑛 - 1), 𝑒) = (𝑛 𝑛 - 1, 𝑒) = (𝑒, 𝑒) (𝜑 𝑎 - 1 (𝑛 - 1), 𝑎 - 1) • (𝑛, 𝑎) = (𝜑 𝑎 - 1 (𝑛 - 1) 𝜑 𝑎 - 1 (𝑛), 𝑎 - 1 𝑎) = (𝜑 𝑎 - 1 (𝑛 - 1 𝑛), 𝑒) = (𝜑 𝑎 - 1 (𝑒), 𝑒) = (𝑒, 𝑒)

as required.

Right action convention

If we instead have right actions, we define the product by $𝑁 ⋊ 𝜑 𝐻$

(𝑛 1, ℎ 1) (𝑛 2, ℎ 2) = (𝑛 1 𝑛 (ℎ - 1 1) 𝜑 2, ℎ 1 ℎ 2)

for $𝑛 𝑖 \in 𝑁$ and $ℎ 𝑖 \in 𝐻$ with

(𝑛, ℎ) - 1 = ((𝑛 - 1) (ℎ) 𝜑, ℎ - 1)

for $𝑛 \in 𝑁$ , $ℎ \in 𝐻$ .

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 $𝑁 ⋊ 𝜑 𝐴$ , #m/thm/group where $𝜑$ denotes the conjugation action (which leave $𝑁$ invariant by normality).

Likewise, if $𝐺$ is the external semidirect product $𝑁 ⋊ 𝜑 𝐴$ , then

the subset $𝑁 𝐺 = 𝑁 \times {𝑒 𝐴} \subseteq 𝐺$ is a normal subgroup isomorphic to $𝑁$
the subset $𝐴 𝐺 = {𝑒 𝑁} \times 𝐴 \subseteq 𝐺$ is a subgroup isomorphic to $𝐴$
$𝐺$ is the internal semidirect product $𝑁 𝐺 ⋊ 𝐴 𝐺$
Conjugation of an element of $𝑁 𝐺$ by an element of $𝐴 𝐺$ 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 $𝐴 \subseteq 𝐺$ be subgroups such that $𝐺 = 𝑁 𝐴$ and $𝑁 \cap 𝐴 = {𝑒}$ . Since $𝑁$ is normal it is invariant under $I n n (𝐺)$ , so conjugation by elements of $𝐴$ is a group action on $𝑁$ , which we denote $𝜑$ . Let $˜ 𝐺 = 𝑁 ⋊ 𝜑 𝐴$ , and $Ψ : ˜ 𝐺 \to 𝐺 : (𝑛, 𝑎) \mapsto 𝑛 𝑎$ . Then for any $𝑛, 𝑚 \in 𝑁$ and $𝑎, 𝑏 \in 𝐴$

Φ ((𝑛, 𝑎) • (𝑚, 𝑏)) = Φ (𝑛 𝜑 𝑎 𝑚, 𝑎 𝑏) = Φ (𝑛 𝑎 𝑚 𝑎 - 1, 𝑎 𝑏) = 𝑛 𝑎 𝑚 𝑎 - 1 𝑎 𝑏 = 𝑛 𝑎 𝑚 𝑏 = Φ (𝑛, 𝑎) Φ (𝑚, 𝑏)

hence $Φ$ is a homomorphism, in particular it is an epimorphism $𝑁 𝐴 = 𝐺$ . Now let $𝑛 \in 𝑁$ and $𝑎 \in 𝐴$ such that $Φ (𝑛, 𝑎) = 𝑛 𝑎 = 𝑒$ . 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, $𝜑 - : 𝐴 \to A u t (𝑁)$ be a (left) group action of $𝐴$ on $𝑁$ , and $𝐺 = 𝑁 ⋊ 𝜑 𝐴$ . Furthermore let $𝑁 𝐺 = 𝑁 \times {𝑒}$ and $𝐴 𝐺 = {𝑒} \times 𝐴$ as sets. Since $(𝑛, 𝑒) • (𝑚, 𝑒) = (𝑛 𝑚, 𝑒)$ for any $𝑛, 𝑚 \in 𝑁$ , and likewise $(𝑒, 𝑎) • (𝑒, 𝑏) = (𝑒, 𝑎 𝑏)$ , it is clear that $𝑁 𝐺$ and $𝐴 𝐺$ are subgroups of $𝐺$ isomorphic to $𝑁$ and $𝐴$ respectively.

Note that

𝑁 𝐺 𝐴 𝐺 = (𝑁, 𝑒) • (𝑒, 𝐴) = (𝜑 𝐴 𝑁, 𝐴) = (𝑁, 𝐴)

so $𝐺 = 𝑁 𝐺 𝐴 𝐺$ .

Let $𝑛 \in 𝑁$ and $𝑎 \in 𝐴$ . Then

(𝑒, 𝑎) • (𝑛, 𝑒) • (𝑒, 𝑎) - 1 = (𝑒, 𝑎) • (𝑛, 𝑒) • (𝑒, 𝑎 - 1) = (𝜑 𝑎 (𝑛), 𝑎) • (𝑒, 𝑎 - 1) = (𝜑 𝑎 (𝑛), 𝑒)

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

#state/tidy | #lang/en | #SemBr

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