Direct and Semidirect Products

We develop the two fundamental ways of building new groups from old: the direct product and the semidirect product. After proving the equivalence of internal and external direct products, we define the semidirect product and illustrate it with the dihedral groups, symmetric groups, and groups of order pq.

Folio Official

March 1, 2026

1 The Direct Product of Groups

Definition 1 (External direct product).

Given groups

G

and

H

, the external direct product

G \times H

is the set of ordered pairs

{(g, h) ∣ g \in G, h \in H}

equipped with the componentwise operation

(g_{1}, h_{1}) (g_{2}, h_{2}) = (g_{1} g_{2}, h_{1} h_{2}) .

Proposition 2.

G \times H

is a group. The identity is

(e_{G}, e_{H})

, and the inverse of

(g, h)

(g^{- 1}, h^{- 1})

Proof.

Associativity follows from that of

G

and

H

(g_{1}, h_{1}) ((g_{2}, h_{2}) (g_{3}, h_{3})) = (g_{1}, h_{1}) (g_{2} g_{3}, h_{2} h_{3}) = (g_{1} g_{2} g_{3}, h_{1} h_{2} h_{3})

, which equals

((g_{1}, h_{1}) (g_{2}, h_{2})) (g_{3}, h_{3})

. The identity and inverse properties are verified componentwise. □

Proposition 3.

G \times H

, define

\overset{ˉ}{G} = {(g, e_{H}) ∣ g \in G}

and

\overset{ˉ}{H} = {(e_{G}, h) ∣ h \in H}

. Then:

$\overset{ˉ}{G} ⊴ G \times H$ and $\overset{ˉ}{H} ⊴ G \times H$ .
$\overset{ˉ}{G} \cap \overset{ˉ}{H} = {(e_{G}, e_{H})}$ .
$\overset{ˉ}{G} \overset{ˉ}{H} = G \times H$ .
$\overset{ˉ}{G} ≅ G$ and $\overset{ˉ}{H} ≅ H$ .

Proof.

For (1), take any

(g^{'}, h^{'}) \in G \times H

and

(g, e_{H}) \in \overset{ˉ}{G}

. Then

(g^{'}, h^{'}) (g, e_{H}) (g^{'}, h^{'})^{- 1} = (g^{'} g g^{' - 1}, e_{H}) \in \overset{ˉ}{G}

, so

\overset{ˉ}{G} ⊴ G \times H

. The argument for

\overset{ˉ}{H}

is identical. Parts (2), (3), and (4) follow directly from the definitions. □

2 The Internal Direct Product

Definition 4 (Internal direct product).

Let

G

be a group with normal subgroups

N

and

K

. We say

G

is the internal direct product of

N

and

K

, written

G = N \times K

, if the following three conditions hold:

$N ⊴ G$ and $K ⊴ G$ .
$N \cap K = {e}$ .
$N K = G$ .

Theorem 5 (Recognition criterion for direct products).

N

and

K

are normal subgroups of

G

with

N \cap K = {e}

and

N K = G

, then the map

φ : N \times K \to G

defined by

(n, k) \mapsto nk

is a group isomorphism.

Proof.

We first show that every element of

N

commutes with every element of

K

. For

n \in N

and

k \in K

, consider the commutator

nk n^{- 1} k^{- 1}

. Since

K ⊴ G

, we have

nk n^{- 1} \in K

, so

nk n^{- 1} k^{- 1} = (nk n^{- 1}) k^{- 1} \in K

. Since

N ⊴ G

, we have

k n^{- 1} k^{- 1} \in N

, so

nk n^{- 1} k^{- 1} = n (k n^{- 1} k^{- 1}) \in N

. Thus

nk n^{- 1} k^{- 1} \in N \cap K = {e}

, giving

nk = kn

It follows that

φ

is a homomorphism:

φ ((n_{1}, k_{1}) (n_{2}, k_{2})) = φ (n_{1} n_{2}, k_{1} k_{2}) = n_{1} n_{2} k_{1} k_{2} = n_{1} k_{1} n_{2} k_{2} = φ (n_{1}, k_{1}) φ (n_{2}, k_{2}) .

Surjectivity follows from

N K = G

. For injectivity, if

nk = e

then

n = k^{- 1} \in N \cap K = {e}

, so

n = k = e

. □

Corollary 6.

G = N \times K

is an internal direct product, then every element of

N

commutes with every element of

K

nk = kn

for all

n \in N

k \in K

3 Generalized Direct Products

Definition 7.

A group

G

is the internal direct product of normal subgroups

N_{1}, \dots, N_{r}

, written

G = N_{1} \times \dots \times N_{r}

, if:

Each $N_{i} ⊴ G$ .
$N_{i} \cap (N_{1} \dots N_{i - 1} N_{i + 1} \dots N_{r}) = {e}$ for $i = 1, \dots, r$ .
$N_{1} N_{2} \dots N_{r} = G$ .

Example 8.

We have

Z /6 Z ≅ Z /2 Z \times Z /3 Z

. Indeed,

N = {0, 3} ≅ Z /2 Z

and

K = {0, 2, 4} ≅ Z /3 Z

are normal subgroups of

Z /6 Z

with

N \cap K = {0}

and

N + K = Z /6 Z

Proposition 9.

g cd (m, n) = 1

, then

Z / mn Z ≅ Z / m Z \times Z / n Z

(the group-theoretic Chinese Remainder Theorem).

4 Motivation for the Semidirect Product

In a direct product $G = N \times K$ , the elements of $N$ and $K$ commute with each other. However, many groups cannot be decomposed in this way.

Example 10.

Consider the dihedral group

D_{n}

(the symmetry group of a regular

n

-gon, of order

2 n

). The rotation subgroup

R = ⟨ r ⟩ ≅ Z / n Z

is normal, and the subgroup

S = {e, s} ≅ Z /2 Z

generated by a reflection satisfies

R \cap S = {e}

and

RS = D_{n}

. But

S

is generally not normal, and

rs \neq = sr

(in fact

sr s^{- 1} = r^{- 1}

This situation motivates the semidirect product, which generalizes the direct product by allowing one factor to act non-trivially on the other.

5 Definition of the Semidirect Product

Definition 11 (External semidirect product).

Let

N

and

H

be groups and

φ : H \to Aut (N)

a homomorphism. The semidirect product of

N

and

H

with respect to

φ

, denoted

N ⋊_{φ} H

, is the set

N \times H

equipped with the operation

(n_{1}, h_{1}) (n_{2}, h_{2}) = (n_{1} \cdot φ (h_{1}) (n_{2}), h_{1} h_{2}) .

Theorem 12.

N ⋊_{φ} H

is a group.

Proof.

Associativity: Using the facts that each

φ (h)

is an automorphism and

φ

is a homomorphism:

((n_{1}, h_{1}) (n_{2}, h_{2})) (n_{3}, h_{3}) = (n_{1} φ (h_{1}) (n_{2}), h_{1} h_{2}) (n_{3}, h_{3}) = (n_{1} φ (h_{1}) (n_{2}) \cdot φ (h_{1} h_{2}) (n_{3}), h_{1} h_{2} h_{3}) = (n_{1} φ (h_{1}) (n_{2}) \cdot φ (h_{1}) (φ (h_{2}) (n_{3})), h_{1} h_{2} h_{3}) = (n_{1} φ (h_{1}) (n_{2} φ (h_{2}) (n_{3})), h_{1} h_{2} h_{3}) = (n_{1}, h_{1}) (n_{2} φ (h_{2}) (n_{3}), h_{2} h_{3}) = (n_{1}, h_{1}) ((n_{2}, h_{2}) (n_{3}, h_{3})) .

Identity:

(e_{N}, e_{H})

serves as the identity, since

(e_{N}, e_{H}) (n, h) = (e_{N} \cdot φ (e_{H}) (n), h) = (n, h)

Inverse:

(n, h)^{- 1} = (φ (h^{- 1}) (n^{- 1}), h^{- 1})

, as one verifies by direct computation. □

Proposition 13.

G = N ⋊_{φ} H

$\overset{ˉ}{N} = {(n, e_{H}) ∣ n \in N} ⊴ G$ and $\overset{ˉ}{N} ≅ N$ .
$\overset{ˉ}{H} = {(e_{N}, h) ∣ h \in H} \leq G$ and $\overset{ˉ}{H} ≅ H$ (but $\overset{ˉ}{H}$ need not be normal).
$\overset{ˉ}{N} \cap \overset{ˉ}{H} = {(e_{N}, e_{H})}$ and $\overset{ˉ}{N} \overset{ˉ}{H} = G$ .
$(e_{N}, h) (n, e_{H}) (e_{N}, h)^{- 1} = (φ (h) (n), e_{H})$ . That is, $\overset{ˉ}{H}$ acts on $\overset{ˉ}{N}$ via $φ$ .

6 Recognition of Internal Semidirect Products

Theorem 14 (Recognition criterion for semidirect products).

Suppose a group

G

satisfies:

$N ⊴ G$ and $H \leq G$ .
$N \cap H = {e}$ .
$N H = G$ .

Define

φ : H \to Aut (N)

φ (h) (n) = hn h^{- 1}

. Then

G ≅ N ⋊_{φ} H

Proof.

Since

N H = G

and

N \cap H = {e}

, every element

g \in G

can be written uniquely as

g = nh

with

n \in N

and

h \in H

. Define

Φ : N ⋊_{φ} H \to G

(n, h) \mapsto nh

This is a homomorphism:

Φ ((n_{1}, h_{1}) (n_{2}, h_{2})) = Φ (n_{1} h_{1} n_{2} h_{1}^{- 1}, h_{1} h_{2}) = n_{1} h_{1} n_{2} h_{1}^{- 1} \cdot h_{1} h_{2} = n_{1} h_{1} n_{2} h_{2} = Φ (n_{1}, h_{1}) Φ (n_{2}, h_{2}) .

Surjectivity follows from

N H = G

, and injectivity from

N \cap H = {e}

. □

7 Examples of Semidirect Products

Example 15 (Dihedral groups).

D_{n} ≅ Z / n Z ⋊_{φ} Z /2 Z

, where

φ (1) (k) = - k

(negation in

Z / n Z

Example 16 (The symmetric group

S_{3}

S_{3} ≅ A_{3} ⋊ ⟨(12)⟩ ≅ Z /3 Z ⋊ Z /2 Z ≅ D_{3}

Example 17 (Non-abelian groups of order

pq

Let

p < q

be primes with

q \equiv 1 (mod p)

. Then

Aut (Z / q Z) ≅ (Z / q Z)^{\times}

has order

q - 1

, which is divisible by

p

, so it contains an element

α

of order

p

. Setting

φ : Z / p Z \to Aut (Z / q Z)

with

φ (1) = α

, the semidirect product

Z / q Z ⋊_{φ} Z / p Z

is a non-abelian group of order

pq

Example 18 (Trivial semidirect product).

When

φ

is the trivial homomorphism (

φ (h) = id_{N}

for all

h \in H

), the semidirect product reduces to the direct product:

N ⋊_{φ} H = N \times H

8 Summary

Direct product $N \times K$ : both $N ⊴ G$ and $K ⊴ G$ , with $N \cap K = {e}$ and $N K = G$ . Elements of $N$ and $K$ commute.
Semidirect product $N ⋊ H$ : $N ⊴ G$ and $H \leq G$ (not necessarily normal), with $N \cap H = {e}$ and $N H = G$ . The subgroup $H$ acts on $N$ by conjugation.
The semidirect product depends on the choice of homomorphism $φ : H \to Aut (N)$ . Different choices of $φ$ generally yield non-isomorphic groups.

The direct and semidirect products are the fundamental methods for assembling groups from known building blocks — or, conversely, for decomposing a group into simpler pieces. They are indispensable tools for understanding group structure.

Group Theory Algebra Textbook Direct Products Semidirect Products

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Direct and Semidirect Products

Folio Official

March 1, 2026

1 The Direct Product of Groups

Definition 1 (External direct product).

Given groups

G

and

H

, the external direct product

G \times H

is the set of ordered pairs

{(g, h) ∣ g \in G, h \in H}

equipped with the componentwise operation

(g_{1}, h_{1}) (g_{2}, h_{2}) = (g_{1} g_{2}, h_{1} h_{2}) .

Proposition 2.

G \times H

is a group. The identity is

(e_{G}, e_{H})

, and the inverse of

(g, h)

(g^{- 1}, h^{- 1})

Proof.

Associativity follows from that of

G

and

H

(g_{1}, h_{1}) ((g_{2}, h_{2}) (g_{3}, h_{3})) = (g_{1}, h_{1}) (g_{2} g_{3}, h_{2} h_{3}) = (g_{1} g_{2} g_{3}, h_{1} h_{2} h_{3})

, which equals

((g_{1}, h_{1}) (g_{2}, h_{2})) (g_{3}, h_{3})

. The identity and inverse properties are verified componentwise. □

Proposition 3.

G \times H

, define

\overset{ˉ}{G} = {(g, e_{H}) ∣ g \in G}

and

\overset{ˉ}{H} = {(e_{G}, h) ∣ h \in H}

. Then:

$\overset{ˉ}{G} ⊴ G \times H$ and $\overset{ˉ}{H} ⊴ G \times H$ .
$\overset{ˉ}{G} \cap \overset{ˉ}{H} = {(e_{G}, e_{H})}$ .
$\overset{ˉ}{G} \overset{ˉ}{H} = G \times H$ .
$\overset{ˉ}{G} ≅ G$ and $\overset{ˉ}{H} ≅ H$ .

Proof.

For (1), take any

(g^{'}, h^{'}) \in G \times H

and

(g, e_{H}) \in \overset{ˉ}{G}

. Then

(g^{'}, h^{'}) (g, e_{H}) (g^{'}, h^{'})^{- 1} = (g^{'} g g^{' - 1}, e_{H}) \in \overset{ˉ}{G}

, so

\overset{ˉ}{G} ⊴ G \times H

. The argument for

\overset{ˉ}{H}

is identical. Parts (2), (3), and (4) follow directly from the definitions. □

2 The Internal Direct Product

Definition 4 (Internal direct product).

Let

G

be a group with normal subgroups

N

and

K

. We say

G

is the internal direct product of

N

and

K

, written

G = N \times K

, if the following three conditions hold:

$N ⊴ G$ and $K ⊴ G$ .
$N \cap K = {e}$ .
$N K = G$ .

Theorem 5 (Recognition criterion for direct products).

N

and

K

are normal subgroups of

G

with

N \cap K = {e}

and

N K = G

, then the map

φ : N \times K \to G

defined by

(n, k) \mapsto nk

is a group isomorphism.

Proof.

We first show that every element of

N

commutes with every element of

K

. For

n \in N

and

k \in K

, consider the commutator

nk n^{- 1} k^{- 1}

. Since

K ⊴ G

, we have

nk n^{- 1} \in K

, so

nk n^{- 1} k^{- 1} = (nk n^{- 1}) k^{- 1} \in K

. Since

N ⊴ G

, we have

k n^{- 1} k^{- 1} \in N

, so

nk n^{- 1} k^{- 1} = n (k n^{- 1} k^{- 1}) \in N

. Thus

nk n^{- 1} k^{- 1} \in N \cap K = {e}

, giving

nk = kn

It follows that

φ

is a homomorphism:

φ ((n_{1}, k_{1}) (n_{2}, k_{2})) = φ (n_{1} n_{2}, k_{1} k_{2}) = n_{1} n_{2} k_{1} k_{2} = n_{1} k_{1} n_{2} k_{2} = φ (n_{1}, k_{1}) φ (n_{2}, k_{2}) .

Surjectivity follows from

N K = G

. For injectivity, if

nk = e

then

n = k^{- 1} \in N \cap K = {e}

, so

n = k = e

. □

Corollary 6.

G = N \times K

is an internal direct product, then every element of

N

commutes with every element of

K

nk = kn

for all

n \in N

k \in K

3 Generalized Direct Products

Definition 7.

A group

G

is the internal direct product of normal subgroups

N_{1}, \dots, N_{r}

, written

G = N_{1} \times \dots \times N_{r}

, if:

Each $N_{i} ⊴ G$ .
$N_{i} \cap (N_{1} \dots N_{i - 1} N_{i + 1} \dots N_{r}) = {e}$ for $i = 1, \dots, r$ .
$N_{1} N_{2} \dots N_{r} = G$ .

Example 8.

We have

Z /6 Z ≅ Z /2 Z \times Z /3 Z

. Indeed,

N = {0, 3} ≅ Z /2 Z

and

K = {0, 2, 4} ≅ Z /3 Z

are normal subgroups of

Z /6 Z

with

N \cap K = {0}

and

N + K = Z /6 Z

Proposition 9.

g cd (m, n) = 1

, then

Z / mn Z ≅ Z / m Z \times Z / n Z

(the group-theoretic Chinese Remainder Theorem).

4 Motivation for the Semidirect Product

In a direct product $G = N \times K$ , the elements of $N$ and $K$ commute with each other. However, many groups cannot be decomposed in this way.

Example 10.

Consider the dihedral group

D_{n}

(the symmetry group of a regular

n

-gon, of order

2 n

). The rotation subgroup

R = ⟨ r ⟩ ≅ Z / n Z

is normal, and the subgroup

S = {e, s} ≅ Z /2 Z

generated by a reflection satisfies

R \cap S = {e}

and

RS = D_{n}

. But

S

is generally not normal, and

rs \neq = sr

(in fact

sr s^{- 1} = r^{- 1}

This situation motivates the semidirect product, which generalizes the direct product by allowing one factor to act non-trivially on the other.

5 Definition of the Semidirect Product

Definition 11 (External semidirect product).

Let

N

and

H

be groups and

φ : H \to Aut (N)

a homomorphism. The semidirect product of

N

and

H

with respect to

φ

, denoted

N ⋊_{φ} H

, is the set

N \times H

equipped with the operation

(n_{1}, h_{1}) (n_{2}, h_{2}) = (n_{1} \cdot φ (h_{1}) (n_{2}), h_{1} h_{2}) .

Theorem 12.

N ⋊_{φ} H

is a group.

Proof.

Associativity: Using the facts that each

φ (h)

is an automorphism and

φ

is a homomorphism:

((n_{1}, h_{1}) (n_{2}, h_{2})) (n_{3}, h_{3}) = (n_{1} φ (h_{1}) (n_{2}), h_{1} h_{2}) (n_{3}, h_{3}) = (n_{1} φ (h_{1}) (n_{2}) \cdot φ (h_{1} h_{2}) (n_{3}), h_{1} h_{2} h_{3}) = (n_{1} φ (h_{1}) (n_{2}) \cdot φ (h_{1}) (φ (h_{2}) (n_{3})), h_{1} h_{2} h_{3}) = (n_{1} φ (h_{1}) (n_{2} φ (h_{2}) (n_{3})), h_{1} h_{2} h_{3}) = (n_{1}, h_{1}) (n_{2} φ (h_{2}) (n_{3}), h_{2} h_{3}) = (n_{1}, h_{1}) ((n_{2}, h_{2}) (n_{3}, h_{3})) .

Identity:

(e_{N}, e_{H})

serves as the identity, since

(e_{N}, e_{H}) (n, h) = (e_{N} \cdot φ (e_{H}) (n), h) = (n, h)

Inverse:

(n, h)^{- 1} = (φ (h^{- 1}) (n^{- 1}), h^{- 1})

, as one verifies by direct computation. □

Proposition 13.

G = N ⋊_{φ} H

$\overset{ˉ}{N} = {(n, e_{H}) ∣ n \in N} ⊴ G$ and $\overset{ˉ}{N} ≅ N$ .
$\overset{ˉ}{H} = {(e_{N}, h) ∣ h \in H} \leq G$ and $\overset{ˉ}{H} ≅ H$ (but $\overset{ˉ}{H}$ need not be normal).
$\overset{ˉ}{N} \cap \overset{ˉ}{H} = {(e_{N}, e_{H})}$ and $\overset{ˉ}{N} \overset{ˉ}{H} = G$ .
$(e_{N}, h) (n, e_{H}) (e_{N}, h)^{- 1} = (φ (h) (n), e_{H})$ . That is, $\overset{ˉ}{H}$ acts on $\overset{ˉ}{N}$ via $φ$ .

6 Recognition of Internal Semidirect Products

Theorem 14 (Recognition criterion for semidirect products).

Suppose a group

G

satisfies:

$N ⊴ G$ and $H \leq G$ .
$N \cap H = {e}$ .
$N H = G$ .

Define

φ : H \to Aut (N)

φ (h) (n) = hn h^{- 1}

. Then

G ≅ N ⋊_{φ} H

Proof.

Since

N H = G

and

N \cap H = {e}

, every element

g \in G

can be written uniquely as

g = nh

with

n \in N

and

h \in H

. Define

Φ : N ⋊_{φ} H \to G

(n, h) \mapsto nh

This is a homomorphism:

Φ ((n_{1}, h_{1}) (n_{2}, h_{2})) = Φ (n_{1} h_{1} n_{2} h_{1}^{- 1}, h_{1} h_{2}) = n_{1} h_{1} n_{2} h_{1}^{- 1} \cdot h_{1} h_{2} = n_{1} h_{1} n_{2} h_{2} = Φ (n_{1}, h_{1}) Φ (n_{2}, h_{2}) .

Surjectivity follows from

N H = G

, and injectivity from

N \cap H = {e}

. □

7 Examples of Semidirect Products

Example 15 (Dihedral groups).

D_{n} ≅ Z / n Z ⋊_{φ} Z /2 Z

, where

φ (1) (k) = - k

(negation in

Z / n Z

Example 16 (The symmetric group

S_{3}

S_{3} ≅ A_{3} ⋊ ⟨(12)⟩ ≅ Z /3 Z ⋊ Z /2 Z ≅ D_{3}

Example 17 (Non-abelian groups of order

pq

Let

p < q

be primes with

q \equiv 1 (mod p)

. Then

Aut (Z / q Z) ≅ (Z / q Z)^{\times}

has order

q - 1

, which is divisible by

p

, so it contains an element

α

of order

p

. Setting

φ : Z / p Z \to Aut (Z / q Z)

with

φ (1) = α

, the semidirect product

Z / q Z ⋊_{φ} Z / p Z

is a non-abelian group of order

pq

Example 18 (Trivial semidirect product).

When

φ

is the trivial homomorphism (

φ (h) = id_{N}

for all

h \in H

), the semidirect product reduces to the direct product:

N ⋊_{φ} H = N \times H

8 Summary

Direct product $N \times K$ : both $N ⊴ G$ and $K ⊴ G$ , with $N \cap K = {e}$ and $N K = G$ . Elements of $N$ and $K$ commute.
Semidirect product $N ⋊ H$ : $N ⊴ G$ and $H \leq G$ (not necessarily normal), with $N \cap H = {e}$ and $N H = G$ . The subgroup $H$ acts on $N$ by conjugation.
The semidirect product depends on the choice of homomorphism $φ : H \to Aut (N)$ . Different choices of $φ$ generally yield non-isomorphic groups.

Group Theory Algebra Textbook Direct Products Semidirect Products

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Direct and Semidirect Products

1 The Direct Product of Groups

2 The Internal Direct Product

3 Generalized Direct Products

4 Motivation for the Semidirect Product

5 Definition of the Semidirect Product

6 Recognition of Internal Semidirect Products

7 Examples of Semidirect Products

8 Summary

Share your expertise with the world

More from Folio Official

Cosets and Lagrange's Theorem

Composition Series and the Jordan--Hölder Theorem

The Homomorphism Theorems

Group Actions and the Class Equation

Direct and Semidirect Products

1 The Direct Product of Groups

2 The Internal Direct Product

3 Generalized Direct Products

4 Motivation for the Semidirect Product

5 Definition of the Semidirect Product

6 Recognition of Internal Semidirect Products

7 Examples of Semidirect Products

8 Summary

Share your expertise with the world

More from Folio Official

Cosets and Lagrange's Theorem

Composition Series and the Jordan--Hölder Theorem

The Homomorphism Theorems

Group Actions and the Class Equation