# Semidirect products and extensions

**Prerequisites.** Group actions, as described in, e.g., my previous blog post. We also make use of generators and relations.

## 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 \(G\) is a finitely generated abelian group, then we can find prime numbers \(p_1,\ldots,p_k\) and positive integers \(n,n_1,\ldots,n_k\) such that $$G \cong \mathbb{Z}^n \times \mathbb{Z}/p_1^{n_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_k^{n_k}\mathbb{Z};$$ here \(n = 0\) if and only if \(G\) 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 \(N\) and \(K\) be groups. Given a homomorphism \(\varphi:K \to \operatorname{Aut}(N)\), we define the

*external semidirect product of \(N\) by \(K\) with respect to \(\varphi\)*to be the cartesian product \(N \times K\) equipped with the group operation $$(n_1,k_1)(n_2,k_2) = (n_1\varphi_{k_1}(n_2), k_1k_2)$$ for all \(n_1,n_2 \in N\) and \(k_1,k_2 \in K\). We denote the resulting group by \(N \rtimes_\varphi K\).

**Exercise 2.9.** Show that every external semidirect product (N \rtimes_\varphi K) is a group.

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, 1872). Let \(G\) be a finite group of order \(p^nm\), where \(p\) is a prime number and \(m\) is an integer relatively prime to \(p\). The following holds:

**First Sylow theorem.**\(G\) contains a Sylow \(p\)-subgroup of order \(p^n\).**Second Sylow theorem.**All Sylow \(p\)-subgroups are conjugates of one another. In other words, if \(P\) and \(Q\) are Sylow \(p\)-subgroups of \(G\), then there exists a \(g \in G\) such that \(gPg^{-1} = Q\). This, in particular, implies that all Sylow \(p\)-subgroups are isomorphic to one another**Third Sylow theorem.**Let \(n_p\) denote the number of Sylow \(p\)-subgroups of \(G\). We have the following combinatorial restrictions on \(n_p\):- \(n_p\) divides \(\vert G \vert\);
- \(n_p \equiv 1 \mbox{ mod } p\).

We also make use of Lagrange’s theorem and its corollaries repeatedly; let us record a few here:

**Theorem 3.2**(Lagrange). Let \(G\) be a finite group, \(H\) a subgroup of \(G\), and \(g\) an element of \(G\). The following holds:

- \(\vert g \vert\) divides \(\vert G \vert\);
- \(\vert G \vert = \vert H \vert [G:H]\);
- If \(H\) is normal in \(G\), then \(\vert G \vert = \vert H \vert \vert G/H \vert \);
- If \(\varphi:G \to G’\) is a homomorphism, then \(\vert G \vert = \vert \ker \varphi \vert \, \vert \operatorname{im} \varphi \vert.\)

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 \(p\) and \(q\) be prime numbers such that \(p > q\). Every group of order \(pq\) is isomorphic to either the cyclic group \(\mathbb{Z}/pq\mathbb{Z}\) or a group given by the presentation $$\langle a,b \mid a^q = b^p = 1, aba^{-1} = b^m \rangle,$$ where \(m^q \equiv 1 \mbox{ mod } p\) and \(m \not\equiv 1 \mbox{ and } p\). In particular, if \(q \nmid p-1\), 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 12. 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

or

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 \(A \xrightarrow{f} B \xrightarrow{g} C\) be a sequence of group homomorphisms. The sequence is said to be

*exact*in case \(\operatorname{im} f = \ker g\). A longer sequence $$G_1 \xrightarrow{f_1} G_2 \xrightarrow{f_2} \cdots \xrightarrow{f_{n-1}} G_n$$ is said to be exact if each joint \(G_{k-1} \xrightarrow{f_{k-1}} G_k \xrightarrow{f_k} G_{k+1}\) is exact.

**Definition 4.4.**A

*short exact sequence of groups*is an exact sequence of groups of the form $$1 \to A \xrightarrow{f} B \xrightarrow{g} C \to 1,$$ where \(1\) denotes the trivial group, \(1 \to A\) denotes the group homomorphism that maps the identity to the identity, and \(C \to 1\) 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 \(G\) is a sequence of subgroups $$G = G_0 \geq G_1 \geq \cdots \geq G_n = 1$$ such that \(G_{k+1} \triangleleft G_k\) for all \(0 \leq k \leq k-1\). The

*\(k\)th factor group*of the above normal series is the quotient group \(G_k/G_{k+1}\). 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 \(G_{k+1}\) is either a maximal proper normal subgroup of \(G_k\) or \(G_{k+1} = G_k\).

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 $$1 \to N \xrightarrow{f} G \xrightarrow{g} K \to 1$$ is

*left split*if there exists a group homomorphism \(G \xrightarrow{f’} N\) such that \(f’ \circ f = \operatorname{id}_N\). Similarly, the short exact sequence is said to be

*right split*if there exists a group homomorphism \(K \xrightarrow{g’} G\) such that \(g \circ g’ = \operatorname{id}_K\). Show that a group extension \(G\) of \(N\) by \(K\) is a semidirect product of \(N\) by \(K\) if and only if the corresponding short exact sequence is right split. Show also that a group extension \(G\) of \(N\) by \(K\) is a direct product of \(N\) and \(K\) 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, Weibel, *An Introduction to Homological
Algebra*.