There are two ways to think of the construction. One is intrinsic: the condition that a given group is a semidirect product of two given subgroups and is equivalent to some special conditions on the subgroups. This concept is usually called an inner semidirect product: Expressing a group as an inner semidirect product of two subgroups can help in the study of the group's behavior. It can also help classify up to isomorphism.
Another way to think of semidirect products is extrinsic: given two abstract groups with some specified relationship between them, given by a certain homomorphism (as defined below), one can construct a new group called the semidirect product (or outer semidirect product) In this way, one can build new, larger groups from smaller ones, with a construction that is more general and richer than a direct product.
The following theorem gives three equivalent definitions of an inner semidirect product.
Let be a group and let and be subgroups of with normal. Then the following statements are equivalent:
Every can be written uniquely as with
Define in the natural way: Then is an isomorphism.
If these conditions hold, one writes and says that this expresses as an inner semidirect product of and
Note that must be normal for these statements to make sense.
The proof of the theorem is straightforward: see the last section of the wiki.
Let Let be the normal subgroup of order 3 generated by a 3-cycle, and let be a subgroup of order 2 generated by a 2-cycle. Then (e.g. by condition (1)).
This example generalizes along two different lines: first, let a subgroup of order 2 generated by a 2-cycle. Then in exactly the same way.
Second, let the dihedral group of order Then let be the index-2 subgroup generated by rotations of the -gon, and let be an order-2 subgroup generated by a reflection. Then
In verifying that a group is a semi-direct product of two given subgroups, it is often useful to note that if is finite and then it suffices to check exactly one of the two conditions and (e.g. by the Second Isomorphism Theorem). And further, if and are relatively prime and the condition is automatic by Lagrange's theorem, so both conditions hold automatically.
Let be a group of order for distinct primes Let be its Sylow -subgroup and suppose that is normal. Show that for some subgroup
Let be a Sylow -subgroup. Then and and are coprime, so the above remarks show immediately that
The outer semidirect product is a sort of generalization of the inner semidirect product. Instead of starting with and prescribed subgroups and the outer semidirect product starts only with the abstract subgroups and constructs the semidirect product The construction does require a notion of multiplication of elements of with elements of ; this extra structure is part of the data that is used to construct the group
Let and be groups, and suppose is a homomorphism, which sends elements to automorphisms of Then the group is defined as the set of ordered pairs with and group law given by the formula
The verification that is a group is nontrivial but straightforward.
Note that the group depends very strongly on the homomorphism Leaving and unchanged but changing will change the group (even up to isomorphism--two different can lead to two non-isomorphic groups ).
Note that this group law is a sort of "twisted" version of the group law of the direct product. Another way to say this is that direct products are trivial examples of semidirect products:
If and are any groups, and is the trivial homomorphism, sending every to the identity automorphism of then
An outer semidirect product is an inner semidirect product. That is, given and a homomorphism the group has a normal subgroup and a subgroup such that (Of course and )
An inner semidirect product is an outer semidirect product. That is, given a group with subgroups and such that there is a natural homomorphism such that That homomorphism is as follows: for define Then which looks like the group law for an outer semidirect product; in other words, the above computation shows that the bijection defined by is also a homomorphism.
Let and be groups, let be a homomorphism, and let be the (outer) semidirect product. Suppose that is abelian. Which of the following three statements must necessarily be true about ?
I. and are abelian
II. is the trivial homomorphism
III. is actually a direct product of groups
Let and let Consider the automorphism which sends the nontrivial element of to the homomorphism Then the group is a group of order 8. Which group of order 8 is it isomorphic to?
The most illuminating way to show that is isomorphic to the dihedral group is to find subgroups isomorphic to and in such that the conjugation action of on is the same as the effect of Let be a 90-degree rotation and a reflection. Then and
Let and let Then and as desired. What does conjugation by the nontrivial element of do to ? It sends to and vice versa, and fixes and Picking the isomorphism that sends and to and respectively, we see that conjugation by has the effect of switching and and preserving and which is precisely the automorphism
Note that this gives an example of the phenomenon mentioned above, that different choices of can lead to nonisomorphic groups: if is chosen to be the trivial homomorphism, then
Let with group operation given by matrix multiplication. Let be the subgroup of consisting of diagonal matrices and let be the subgroup of consisting of matrices of the form Then is normal, and so
Abstractly, and , so is isomorphic to the (outer) semidirect product where sends a nonzero real number to the automorphism of the additive group given by multiplication by
Another way to view is as a group of affine transformations Then matrix multiplication in corresponds to composition of the associated affine transformations.
Since every can be written as The representation is unique because, if then The left side is in and the right side is in ; since they must both be so and
To see that is surjective, note that any coset can be rewritten using (2) as for some (This is because because is normal.) To see that is injective, suppose Then but then expresses in two ways as a product of an element in and an element in This representation is supposed to be unique, so must equal
If then is the identity of so because is injective. Now for any for some because is surjective. So so for some so This shows that