1. Introduction: Direct Products
Recall that the external direct product of two groups and is the cartesian product equipped with coordinatewise group operation: for all and . The external direct product of and is arguably the simplest and the most natural way to construct a new group out of and . Therefore, it is of interest to determine whether a group can be represented as a direct product of its subgroups.
Example 1.1. The integers mod 6 group is isomorphic to . Indeed, the map
given by the formula is a group isomorphism.
Example 1.2. The Klein-four group is isomorphic to . Indeed, the map
given by the formula is a group isomorphism.
Example 1.3. The multiplicative group of complex numbers , equipped with the standard multiplication operation, is isomorphic to the direct product of the circle group and the group of positive real numbers , both equipped with the standard multiplication operation. Indeed, every nonzero complex number can be written uniquely as , and so the map
given by the formula is a group isomorphism.
The question, then, is as follows: given two subgroups and of a group , when is isomorphic to ? The above examples suggest that we should be able to create an arbitrary element of by taking the product of an element of and an element of . Moreover, in all of the examples above, and are effectively disjoint, in the sense that . These observations lead us to the following definition:
Definition 1.4. Two subgroups and of are permutable complements—or simply complements if there is no danger of confusion—in if and equals .
Is this sufficient? Not so, as the next example shows.
Example 1.5. Consider the dihedral group
Let and , so that . Since every element of is of the form , we see that . Nevertheless, cannot be isomorphic to : is nonabelian, but is abelian.
What additional conditions do we need?
Proposition 1.6. Let and be subgroups of a group . If and are permutable complements in , , and , then .
Proof. Define a mapping by setting . Since , the mapping is surjective. If , then , so that
. Since the intersection is trivial, we see that and . It follows that is injective.
It remains to show that is a group homomorphism. To this end, we show that whenever and . Indeed, the normality of in implies that , and the normality of in implies that . Since the intersection is trivial, we see that , and so .
We now observe that
for all and , whence is a homomorphism.
The converse also holds.
Proposition 1.7. If , then has normal subgroups and such that , , and and are permutable complements in .
Proof. Let and . We see at once that and are permutable complements in . Moreover, it follows from the definition of the direct product that whenever and .
Fix . For each , we can find and such that . Therefore,
by the commutativity of elements of and elements of . Since and were arbitrary, we conclude that . Similarly, we can show that .
We can summarize the above two propositions as the equivalence of external direct product and internal direct product.
Definition 1.8. A group is said to be the internal direct product of subgroups and if and are permutable complements in , , and .
As an application, we show that can be written as a direct product of two subgroups.
Example 1.9. If is odd, then the general linear group is isomorphic to the direct product of the group of scalar matrices and the special linear group . Given , we let , so that is of determinant 1. We can then write as the product of a scalar matrix and a matrix of determinant 1. Since was arbitrary, we see that .
Now, is odd, and so a scalar matrix is of determinant 1 if and only if . Therefore, , and it follows that and are permutable complements in .
Since scalar matrices commute with all matrices of compatible size, . Moreover, if and , then
and so . It follows that . We conclude from Proposition 1.6 that .
An important classification theorem regarding direct products is the following:
Theorem 1.10 (Fundamental theorem on finitely generated abelian groups). If is a finitely generated abelian group, then we can find prime numbers and positive integers such that
here if and only if is finite.
2. Semidirect Products
Classification results with direct products, even something as strong as Theorem 1.10, does not take into account the decomposition of into the “internal product” that was discussed in Example 1.5. It is therefore reasonable to generalize the notion of internal direct products introduced in Definition 1.8 for the sake of completeness.
Definition 2.1. Let be a group, and fix subgroups and of . The group is said to be an internal semidirect product of by , denoted by , if and and are permutable complements of .
The notation emphasizes the fact that the left component, , is the normal subgroup. Since Definition 2.1 is a straightforward generalization of Definition 1.8, we see that every internal direct product is an internal semidirect product. This, in particular, implies that internal semidirect products are not necessarily isomorphic.
Example 2.2. Consider the dihedral group and the integers mod 6 group . By Example 1.1 and Example 1.8, each group is an internal semidirect product of a subgroup isomorphic to by a subgroup isomorphic to . Nevertheless, two groups fail to be isomorphic: is nonabelian, whereas is abelian.
Before we consider more examples of internal semidirect products, we make the following observation: the two “parts” of a semidirect product do not necessarily commute. In the proof of Proposition 1.6, we have shown that if is an internal direct product of and , then whenever and . Nevertheless, we know that in , and so the same principle fails to hold for semidirect products. What we do have, however, is the following:
Proposition 2.3. Let be a group, a normal subgroup of , and a subgroup of . Then is a subgroup of and .
Proof. Given arbitrary , we observe that
for some by the normality of in . It follows that . Similarly, . Now, both and coincide with the subgroup of generated by , and so .
Let us now consider two examples of internal semidirect products, and one counterexample.
Example 2.4. Fix . The symmetric group is an internal semidirect product of by . Indeed, because conjugation preserves parity. Moreover, every transposition is in , as . Note also that contains all odd permutations. Since every permutation is a product of transpositions, it follows from the subgroup property established in Proposition 2.3 that . Finally, every permutation in is even, but is an odd permutation, and so and are permutable complements. It now follows that is an internal semidirect product of by .
Example 2.5. Generalizing Example 1.5. and Example 2.2., we consider the dihedral group
Let and , so that and . Observe now that every element of is of the form . Given and , we see that
It follows that , whence is an internal semidirect product of by .
Example 2.6. Consider the unit quaternion group , introduced in Example 2.8 of my blog post on generators and relations. Since every nontrivial subgroup of must contain , it follows that no two nontrivial subgroups of are permutable complements in . Therefore, is not an internal semidirect product of its nontrivial subgroups.
We now turn to the problem of constructing an external semidirect product of two groups, without considering them as subgroups of an ambient group. Note that if , then the conjugation action of on , given by the formula , is a well-defined group homomorphism by the normality of in . In this case, we see that the generic group product in takes the following form:
Taking a cue from this computation, we make the following definition:
Definition 2.8. Let and be groups. Given a homomorphism , we define the external semidirect product of by with respect to to be the cartesian product equipped with the group operation
for all and , which we denote by .
We show that the construction we have given above yields a group.
Exercise 2.9. Every external semidirect product is a group. (This was given as a homework exercise in class.)
Mimicking Proposition 1.6, we show that an internal semidirect product is isomorphic to the corresponding external semidirect product.
Proposition 2.10. Let and be subgroups of a group . If and are permutable complements in , , and , then , where is the conjugation map .
Proof. We define a map by setting . Observe that
by what we have shown in (2.7), and so is a group homomorphism. If , then , whence . Since the intersection is trivial, we see that , whence the kernel of is trivial. Finally, is surjective, as .
Conversely, an analogue of Proposition 1.7 holds as well.
Proposition 2.11. If , then has a normal subgroup and a subgroup such that , , and and are permutable complements in .
Proof. Let and , so that and are permutable complements in . To show that , we fix and . We first compute . To this end, we let and observe that
Therefore, . On the other hand,
so that . Therefore
We now observe that
It follows that , as was to be shown.
Therefore, the knowledge of automorphism groups is paramount in determining whether a group can be realized as a semidirect product. Here is a simple example; see Section 7.2 of Rotman, An Introduction to the Theory of Groups for a more comprehensive survey.
Example 2.12. Let and be finite groups, and let be a group homomorphism. Lagrange’s theorem, in conjunction with the first isomorphism theorem, implies that , and so divides . Moreover, Lagrange’s theorem implies that divides . Therefore, if and are relatively prime, then must be the trivial homomorphism, and so must be the direct product .
Now, we observe that whenever is prime. Indeed, Lagrange’s theorem implies that every non-identity element of is a generator. Now, every map from a cyclic group to a cyclic group that maps a generator to a generator can be extended to a group homomorphism (see my blog post on homomorphic extensions. Since there are generators, there are such homomorphisms from into itself. It is easy to show that all such maps are automorphisms, and that no other map is an automorphism. The isomorphism statement now follows.
Therefore, there is no non-direct semidirect product , as 5 and are relatively prime. On the other hand, there is precisely one non-direct semidirect product , as there are precisely two ways of mapping homomorphically into : trivially or surjectively.
3. Classification of Groups of Small Order
As an application of the product constructions discussed so far, we establish several classifaction results for finite groups. To this end, we shall make use of the Sylow theorems, which was covered in my blog post on group actions Let us state the theorems once again, for ease of reference:
Theorem 3.1 (Sylow). Let be a finite group of order , where is a prime number and is an integer relatively prime to . The following holds:
First Sylow theorem. contains a Sylow -subgroup of order .
Second Sylow theorem. All Sylow -subgroups are conjugates of one another. In other words, if and are Sylow -subgroups of , then there exists a such that . This, in particular, implies that all Sylow -subgroups are isomorphic to one another.
Third Sylow theorem. Let denote the number of Sylow -subgroups of . We have the following combinatorial restrictions on :
- divides ;
We also make use of Lagrange’s theorem and its corollaries repeatedly; let us record a few here:
Theorem 3.2 (Lagrange). Let be a finite group, a subgroup of , and an element of . The following holds:
- divides ;
- If is normal in , then ;
- If is a homomorphism, then
Finally, we also rely on the fundamental theorem on finitely generated abelian groups (Theorem 1.10). Since we have not proved the theorem in class or in the notes, you will not be responsible for any classification result that makes use of the theorem; all such instances are marked in red.
The first result is a trivial corollary of Lagrange’s theorem
Proposition 3.2. Every finite group of prime order is cyclic.
Proof. Let be a group of order . By Lagrange’s theorem, every non-identity element of is of order . It follows that is cyclic.
The second result makes use of the fundamental theorem on finitely generated abelian groups.
Theorem 3.4. Let be a prime number. Every group of order is isomorphic to either or .
Proof. We shall make use of two lemmas:
Lemma 3.5. If is a -group, then .
Proof of lemma. Consider the conjugacy class equation (Theorem 3.11 in my blog post on group actions):
Since for all , Lagrange’s theorem implies that each is divisible by . Since is divisible by , it follows that is divisible by .
Lemma 3.6. If is cyclic, then is abelian.
Proof of lemma. Let be the canonical projection map, and find such that is a generator of . Fix and find such that and . We can find such that and . We now observe that
Since and were arbitrary, we conclude that is abelian.
We now let be a group of order . Suppose for a contradiction that is nonabelian. By Lagrange’s theorem, is either 1 or . Lemma 3.5 implies that . Since is normal in , we apply Lagrange’s theorem to conclude that . It follows from Proposition 3.2 that is cyclic. Lemma 3.6 now implies that is abelian, which is evidently absurd. It follows that is abelian, whence the fundamental theorem on finitely generated abelian groups (Theorem 1.10) implies the desired classification result.
The next result is another one of general nature, covering the two-factor cases that Theorem 3.4 leaves out.
Theorem 3.7. Let and be prime numbers such that . Every group of order is isomorphic to either the cyclic group or a group given by the presentation (see my blog post on generators and relations).
where and . In particular, if , then the second case cannot occur.
Proof. We shall make use of three lemmas:
Lemma 3.8. Let be a finite group containing a subgroup of index , where is the smallest prime divisor of . Then .
Proof of lemma. Let denote the left coset space , so that . Let be the action of on by left multiplication (see Example 2.12 in my blog post on group actions). By the first isomorphism theorem, the quotient group is isomorphic to a subgroup of . Since , Lagrange’s theorem implies that divides .
Now, Lagrange’s theorem also implies that divides . Since is the smallest prime that divides , we must have . The action is nontrivial, and so we must have .
For each , we see that , and so . Therefore, . Lagrange’s theorem implies that
It follows that , and so . Since is normal in , the proof is complete.
Lemma 3.9. Let be a finite group of order . If all of the Sylow subgroups of are normal, then isomorphic to the direct product of its Sylow subgroups.
Proof of lemma. The third Sylow theorem, in conjunction with the second Sylow theorem, implies that has precisely one Sylow -subgroup and one Sylow -subgroup . By Lagrange’s theorem, . Moreover, , an dso . Therefore, and are permutable complements in , both of which are normal in by hypothesis. It follows from Proposition 1.6 that .
Lemma 3.10 (Chinese remainder theorem for groups). If and are relatively prime positive integers, then .
Proof of lemma. Consider , which is of order . Let and , so that and . Since is abelian, we see that for all . As and are relatively prime, , and the product group is cyclic.
We now let be a group of order . The first Sylow theorem implies that contains a Sylow -subgroup . Since |P| = p, Proposition 3.2 implies that is cyclic; let be an element of of order . Lemma 3.8 implies that .
The first Sylow theorem also implies that contains a Sylow -subgroup . Once again, we can invoke Proposition 3.2 to find an element of of order . If , then the second Sylow theorem implies that , whence by Lemma 3.9. Proposition 3.2 implies that and , and it follows from Lemma 3.10 that .
We therefore assume that . The third Sylow theorem furnishes a positive integer such that . The third Sylow theorem also implies that divides , whence we must have . This, in particular, implies that , and so .
Now, , and so for some . If , then , and is abelian. Let us therefore assume that .
We claim that for all . This, in particular, implies that , whence , as was to be shown. It therefore suffices to establish the claim.
We proceed by induction. The case has already been established. We fix and assume that the case holds. Then
as was to be shown.
Remark 3.11. By generalizing the notion of internal direct products to more than two subgroups, we can establish the following extension of Lemma 3.9: If is a finite group such that all of its Sylow subgroups are normal, then is isomorphic to the direct product of all of its Sylow subgroups.
Corollary 3.12. If is prime, then every group of order is either cyclic or isomorphic to the dihedral group .
Proof. If the group is non-cyclic, then it is isomorphic to a group given by the presentation
where and . Among , we see that is the only choice that satisfies both restrictions. It now suffices to note that the above presentation with yields the dihedral group .
The only groups of order at most that are not covered by the results established thus far are groups of order 8 and groups of order
- We now tackle them “by hand”.
Theorem 3.13. Every group of order 8 is isomorphic to , , , the unit quaternion group , or the dihedral group .
Proof. Let be a group of order 8. By Lagrange’s theorem, every element of is of order 1, 2, 4, or 8. If contains an element of order 8, then .
Let us suppose that has no element of order 8. If has no element of order 4, then every non-identity element of is of order 2. Whenever , we have the identity
whence is abelian. Fix three non-identity elements of and define a mapping by setting It is routine to check that is a group isomorphism.
Let us now suppose that has an element of order . Assume for now that is abelian. Fix an element of order 4, and pick an element of order 2 that is not a power of . Define a mapping by setting . Once again, it is easy to check that is a group isomorphism.
Finally, we suppose that is a nonabelian group of order 8 that contains an element of order 4. Since , we see that : see Lemma 4.2 in my blog post on group actions for a proof). In particular, . Therefore, every such that satisfies the property that .
We now fix such a . If or , then is of order 8, which is absurd. We must therefore have or .
Moreover, is normal in , and so for some . cannot be 0, as . cannot be 1, as is nonabelian. cannot be 2, as as and must have the same order. We conclude that .
Therefore, is isomorphic to either
The first case corresponds to ; the second case corresponds to .
Theorem 3.15. Every group of order 12 is isomorphic to , , , , or , where is the unique nontrivial group homomorphism (see Example 2.12).
Proof. The conclusion in red follows from the fundamental theorem on finitely generated abelian groups (Theorem 1.10) and the Chinese remainder theorem for groups (Lemma 3.10). We suppose that is a nonabelian group of order 12 and show that is isomorphic to , , or . To this end, we assume without loss of generality that .
The first Sylow theorem furnishes a Sylow 3-subgroup , which is of order 3. We show that . We can think of each element of as a permutation on the set of all left cosets of ; since , this furnishes a homomorphism such that . , and so Lagrange’s theorem implies that is either 1 or 3. If is trivial, then is isomorphic to a subgroup of of order 12. Since is the only subgroup of of order 12, we see that , which is absurd. Therefore, , and so . It follows that .
The first Sylow theorem also furnishes a Sylow 2-subgroup , which is of order 4. By Lagrange’s theorem , and so . Moreover, , and so . It now follows from Proposition 2.10 that . Since and , there are two possible homomorphisms : the trivial homomorphism, and the surjective homomorphism (see Example 2.12). The first case corresponds to the direct product . Since and are both abelian, this implies that is abelian, which is absurd.
The second case corresponds to the semidirect product , where is the unique surjective homomorphism. If , then . If , then .
4. Exact Sequences and the Extension Problem
We now approach the study of semidirect product from a different angle. We first note that is obtained by piecing together and .
Proposition 4.1. .
Proof. It suffices to observe that
In light of this, we make the following definition:
Definition 4.2. Let and be groups. We say that is an extension of by if has a normal subgroup such that and .
Observe that is an extension of by if and only if there exist an injective homomorphism and a surjective homomorphism such that . Indeed, if such maps exist, then and
Conversely, if we have isomorphisms and , then we can define a mapping by setting and check that .
We isolate the crucial condition:
Definition 4.3. Let be a sequence of group homomorphisms (see my blog post on how to read diagrams). The sequence is said to be exact in case . A longer sequence
is said to be exact if each joint is exact.
Definition 4.4. A short exact sequence of groups is an exact sequence of groups of the form
where denotes the trivial group, denotes the group homomorphism that maps the identity to the identity, and denotes the trivial homomorphism.
We make a few observations. Since the image of is trivial, the exactness condition on means that , or that is injective. The kernel of is , and so the exactness condition on implies that , or that is surjective. Finally, the exactness condition on states merely that . We thus see that short exact sequences are precisely the embodiments of extensions of groups:
Proposition 4.5. is an extension of by if and only if there exists a short exact sequence .
The extension problem, then, is to determine all groups that makes the sequence exact for fixed and . The extension problem lies at the heart of the classification of finite groups. To see why, we need a few notions from the theory of normal series:
Definition 4.6. A normal series of a group is a sequence of subgroups
such that for all . The th factor group of the above normal series is the quotient group . The length of the above normal series is the number of nontrivial factor groups. The above normal series is said to be a composition series in case each is either a maximal proper normal subgroup of or .
The Jordan–Hölder theorem states that each group has a unique composition series in a precise sense. Now, we take a composition series
of and let be the corresponding factor groups. Note that, at each stage, is an extension of by . Moreover, we observe that each factor group is either simple or trivial. Therefore, an arbitrary finite group can be obtained by carrying out the “extending by a finite simple group” process finitely many times—and, by the Jordan–Hölder theorem, this procedure is uniquely determined by .
In other words, we can complete the classification of finite groups by (1) classifying all finite simple groups and (2) solving the extension problem. (1) has been completed, after 60 years of hard work by many group theorists. (2) is still unsolved, as there is no known general theory of classifying all extensions of a given group: see the Wikipedia page on group extensions.
The unresolved state of the group extension problem indicates that there are, in general, many group extensions that fail to be semidirect products. The following exercise sheds light on this phenomenon:
Exercise 4.7. We say that a short exact sequence of groups
is left split if there exists a group homomorphism such that . Similarly, the short exact sequence is said to be right split if there exists a group homomorphism such that . Show that a group extension of by is a semidirect product of by if and only if the corresponding short exact sequence is right split. Show also that a group extension of by is a direct product of and if and only if the corresponding short exact sequence is left split. Conclude that left split implies right split.
See Chapter 7 of Rotman, An Introduction to the Theory of Groups for a more detailed survey of the group extension problem. The extension problem for abelian groups (and, in general, abelian categories) is also heavily studied. The proper context for this type of problem is homological algebra: see, for example, Chapter Weibel, An Introduction to Homological Algebra.