semidirect product of groups

semidirect product of groups

The goal of this exposition is to carefully explain the correspondence
between the notions of external and internal semi–

direct products of
groups

, as well as the

connection

between semi–

direct products

and

short exact sequences

.

Naturally, we start with the construction of semi–direct products.

Definition 1

.

Let

H

and

Q

be groups and let

θ

:

Q

⟶

Aut

⁡

(

H

)

be a

group
homomorphism

. The

semi–direct product

H

⋊

θ

Q

is defined to be the

group with

underlying set

{

(

h

,

q

)

∣

h

∈

H

,

q

∈

Q

}

and

group operation

(

h

,

q

)

⁢

(

h

′

,

q

′

)

:=

(

h

⁢

θ

⁢

(

q

)

⁢

h

′

,

q

⁢

q

′

)

.

We leave it to the reader to check that

H

⋊

θ

Q

is
really a group. It helps to know that the

inverse

of

(

h

,

q

)

is

(

θ

⁢

(

q

-

1

)

⁢

(

h

-

1

)

,

q

-

1

)

.

For the remainder of this article, we omit

θ

from the notation
whenever this map is clear from the

context

.

Set

G

:=

H

⋊

Q

. There exist canonical

monomorphisms

H

⟶

G

and

Q

⟶

G

, given by

h

↦

(

h

,

1

Q

)

,

h

∈

H

q

↦

(

1

H

,

q

)

,

q

∈

Q

where

1

H

(resp.

1

Q

) is the

identity element

of

H

(resp.

Q

). These monomorphisms are so natural that we will treat

H

and

Q

as

subgroups

of

G

under these

inclusions

.

Theorem 2

.

Let

G

:=

H

⋊

Q

as above. Then:

•

H

is a

normal subgroup

of

G

.

•

H

⁢

Q

=

G

.

•

H

∩

Q

=

{

1

G

}

.

Proof.

Let

p

:

G

⟶

Q

be the

projection map

defined by

p

⁢

(

h

,

q

)

=

q

. Then

p

is a

homomorphism

with kernel

H

. Therefore

H

is a normal
subgroup of

G

.

Every

(

h

,

q

)

∈

G

can be written as

(

h

,

1

Q

)

⁢

(

1

H

,

q

)

. Therefore

H

⁢

Q

=

G

.

Finally, it is evident that

(

1

H

,

1

Q

)

is the only element of

G

that is of the form

(

h

,

1

Q

)

for

h

∈

H

and

(

1

H

,

q

)

for

q

∈

Q

.
∎

This result motivates the definition of internal semi–direct
products.

Definition 3

.

Let

G

be a group with subgroups

H

and

Q

. We say

G

is the

internal semi–direct product

of

H

and

Q

if:

•

H

is a normal subgroup of

G

.

•

H

⁢

Q

=

G

.

•

H

∩

Q

=

{

1

G

}

.

We know an external semi–direct product is an internal semi–direct
product (

Theorem

2

). Now we prove a

converse

(Theorem

5

), namely, that an internal semi–direct product is an external semi–direct product.

Lemma 4

.

Let

G

be a group with subgroups

H

and

Q

. Suppose

G

=

H

⁢

Q

and

H

∩

Q

=

{

1

G

}

. Then every

element

g

of

G

can be written
uniquely in the form

h

⁢

q

, for

h

∈

H

and

q

∈

Q

.

Proof.

Since

G

=

H

⁢

Q

, we know that

g

can be written as

h

⁢

q

. Suppose it can
also be written as

h

′

⁢

q

′

. Then

h

⁢

q

=

h

′

⁢

q

′

so

h

′

-

1

⁢

h

=

q

′

⁢

q

-

1

∈

H

∩

Q

=

{

1

G

}

. Therefore

h

=

h

′

and

q

=

q

′

.
∎

Theorem 5

.

Suppose

G

is a group with subgroups

H

and

Q

, and

G

is the
internal semi–direct product of

H

and

Q

. Then

G

≅

H

⋊

θ

Q

where

θ

:

Q

⟶

Aut

⁡

(

H

)

is given by

θ

⁢

(

q

)

⁢

(

h

)

:=

q

⁢

h

⁢

q

-

1

,

q

∈

Q

,

h

∈

H

.

Proof.

By Lemma

4

, every element

g

of

G

can be written
uniquely in the form

h

⁢

q

, with

h

∈

H

and

q

∈

Q

. Therefore,
the map

ϕ

:

H

⋊

Q

⟶

G

given by

ϕ

⁢

(

h

,

q

)

=

h

⁢

q

is a

bijection

from

G

to

H

⋊

Q

. It only remains to show that
this bijection is a homomorphism.

Given elements

(

h

,

q

)

and

(

h

′

,

q

′

)

in

H

⋊

Q

, we have

ϕ

⁢

(

(

h

,

q

)

⁢

(

h

′

,

q

′

)

)

=

ϕ

⁢

(

(

h

⁢

θ

⁢

(

q

)

⁢

(

h

′

)

,

q

⁢

q

′

)

)

=

ϕ

⁢

(

h

⁢

q

⁢

h

′

⁢

q

-

1

,

q

⁢

q

′

)

=

h

⁢

q

⁢

h

′

⁢

q

′

=

ϕ

⁢

(

h

,

q

)

⁢

ϕ

⁢

(

h

′

,

q

′

)

.

Therefore

ϕ

is an

isomorphism

.
∎

Consider the external semi–direct product

G

:=

H

⋊

θ

Q

with subgroups

H

and

Q

. We know from Theorem

5

that

G

is isomorphic to the external semi–direct product

H

⋊

θ

′

Q

, where we are temporarily writing

θ

′

for the

conjugation

map

θ

′

⁢

(

q

)

⁢

(

h

)

:=

q

⁢

h

⁢

q

-

1

of
Theorem

5

. But in fact the two maps

θ

and

θ

′

are the same:

θ

′

⁢

(

q

)

⁢

(

h

)

=

(

1

H

,

q

)

⁢

(

h

,

1

Q

)

⁢

(

1

H

,

q

-

1

)

=

(

θ

⁢

(

q

)

⁢

(

h

)

,

1

Q

)

=

θ

⁢

(

q

)

⁢

(

h

)

.

In summary, one may use Theorems

2

and

5

to pass freely between the notions of internal semi–direct product and external semi–direct product.

Finally, we discuss the correspondence between semi–direct products
and

split exact sequences

of groups.

Definition 6

.

An

exact sequence

of groups

1

⟶

H

⟶

i

G

⟶

j

Q

⟶

1

.

is

split

if there exists a homomorphism

k

:

Q

⟶

G

such that

j

∘

k

is the

identity map

on

Q

.

Theorem 7

.

Let

G

,

H

, and

Q

be groups. Then

G

is isomorphic to a
semi–direct product

H

⋊

Q

if and only if there exists a
split exact sequence

1

⟶

H

⟶

i

G

⟶

j

Q

⟶

1

.

Proof.

First suppose

G

≅

H

⋊

Q

. Let

i

:

H

⟶

G

be the

inclusion map

i

⁢

(

h

)

=

(

h

,

1

Q

)

and let

j

:

G

⟶

Q

be the
projection map

j

⁢

(

h

,

q

)

=

q

. Let the splitting map

k

:

Q

⟶

G

be
the inclusion map

k

⁢

(

q

)

=

(

1

H

,

q

)

. Then the

sequence

above is clearly
split exact.

Now suppose we have the split exact sequence above. Let

k

:

Q

⟶

G

be the splitting map. Then:

•

i

⁢

(

H

)

=

ker

⁡

j

, so

i

⁢

(

H

)

is normal in

G

.

•

For any

g

∈

G

, set

q

:=

k

⁢

(

j

⁢

(

g

)

)

. Then

j

⁢

(

g

⁢

q

-

1

)

=

j

⁢

(

g

)

⁢

j

⁢

(

k

⁢

(

j

⁢

(

g

)

)

)

-

1

=

1

Q

, so

g

⁢

q

-

1

∈

Im

⁡

i

. Set

h

:=

g

⁢

q

-

1

. Then

g

=

h

⁢

q

. Therefore

G

=

i

⁢

(

H

)

⁢

k

⁢

(

Q

)

.

•

Suppose

g

∈

G

is in both

i

⁢

(

H

)

and

k

⁢

(

Q

)

. Write

g

=

k

⁢

(

q

)

. Then

k

⁢

(

q

)

∈

Im

⁡

i

=

ker

⁡

j

, so

q

=

j

⁢

(

k

⁢

(

q

)

)

=

1

Q

. Therefore

g

=

k

⁢

(

q

)

=

k

⁢

(

1

Q

)

=

1

G

, so

i

⁢

(

H

)

∩

k

⁢

(

Q

)

=

{

1

G

}

.

This proves that

G

is the internal semi–direct product of

i

⁢

(

H

)

and

k

⁢

(

Q

)

. These are isomorphic to

H

and

Q

,
respectively. Therefore

G

is isomorphic to a semi–direct product

H

⋊

Q

.
∎

Thus, not all normal subgroups

H

⊂

G

give rise to an (internal)
semi–direct product

G

=

H

⋊

G

/

H

. More specifically, if

H

is a normal subgroup of

G

, we have the canonical exact sequence

1

⟶

H

⟶

G

⟶

G

/

H

⟶

1

.

We see that

G

can be decomposed into

H

⋊

G

/

H

as an internal semi–direct product if and only if the canonical exact sequence splits.

Title

semidirect product of groups

Canonical name

SemidirectProductOfGroups

Date of creation

2013-03-22 12:34:49

Last modified on

2013-03-22 12:34:49

Owner

djao (24)

Last modified by

djao (24)

Numerical id

10

Author

djao (24)

Entry type

Definition

Classification

msc 20E22

Synonym

semidirect product

Synonym

semi-direct product

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