Semidirect product

The semidirect product 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

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 and be subgroups, the first of which is normal, such that and . Then is the internal semidirect product . #m/def/group

External semidirect product

For the external construction, let be a group and let be a group acting on by automorphisms, i.e. equipped with a homomorphism . Then the external semidirect product is the set with group multiplication given by #m/def/group

the identity is , and the inverse is .

Proof of group

For associativity, note

as required. For identity, note

as required. For inverse, note

as required.

Right action convention

If we instead have right actions, we define the product by

for and with

for , .

Relationship between internal and external semidirect product

If is the internal semidirect product , then is isomorphic to the external semidirect product , #m/thm/group where denotes the conjugation action (which leave invariant by normality).

Likewise, if is the external semidirect product , then

the subset is a normal subgroup isomorphic to
the subset is a subgroup isomorphic to
is the internal semidirect product
Conjugation of an element of by an element of is the group action .

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

Proof

Let be a normal subgroup and be subgroups such that and . Since is normal it is invariant under , so conjugation by elements of is a group action on , which we denote . Let , and . Then for any and

hence is a homomorphism, in particular it is an epimorphism . Now let and such that . It follows that , so both and must contain , which can only be true if . Therefore is a monomorphism and thus an isomorphism.

Now let be arbitrary groups, be a (left) group action of on , and . Furthermore let and as sets. Since for any , and likewise , it is clear that and are subgroups of isomorphic to and respectively.

Note that

so .

Let and . Then

as claimed above. This also shows that , since conjugating by any element is equivalent to conjugating by an element of , and then conjugating by an element of . Clearly , so internally.

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

that to which the triangle points, so is normal in and . ↩