1Groups and homomorphisms

IA Groups

1.5 Direct products of groups

Recall that if we have to sets

X, Y

, then we can obtain the product

X × Y

=

{(x, y) : x ∈ X, y ∈ Y }. We can do the same if X and Y are groups.

Definition

(Direct product of groups)

.

Given two groups (

G, ◦

) and (

H, •

),

we can define a set

G × H

=

{

(

g, h

) :

g ∈ G, h ∈ H}

and an operation

(a

1

, a

2

) ∗ (b

1

, b

2

) = (a

1

◦ b

1

, a

2

• b

2

). This forms a group.

Why would we want to take the product of two groups? Suppose we have

two independent triangles. Then the symmetries of this system include, say

rotating the first triangle, rotating the second, or rotating both. The symmetry

group of this combined system would then be D

6

× D

6

.

Example.

C

2

× C

2

= {(0, 0), (0, 1), (1, 0), (1, 1)}

= {e, x, y, xy} with everything order 2

= hx, y | x

2

= y

2

= e, xy = yxi

Proposition. C

n

× C

m

∼

=

C

nm

iff hcf(m, n) = 1.

Proof.

Suppose that

hcf

(

m, n

) = 1. Let

C

n

=

hai

and

C

m

=

hbi

. Let

k

be the

order of (

a, b

). Then (

a, b

)

k

= (

a

k

, b

k

) =

e

. This is possible only if

n | k

and

m | k

, i.e.

k

is a common multiple

n

and

m

. Since the order is the minimum

value of k that satisfies the above equation, k = lcm(n, m) =

nm

hcf(n,m)

= nm.

Now consider

h

(

a, b

)

i ≤ C

n

× C

m

. Since (

a, b

) has order

nm

,

h

(

a, b

)

i

has

nm

elements. Since

C

n

× C

m

also has

nm

elements,

h

(

a, b

)

i

must be the whole of

C

n

× C

m

. And we know that h(a, b)i

∼

=

C

nm

. So C

n

× C

m

∼

=

C

nm

.

On the other hand, suppose

hcf

(

m, n

)

6

= 1. Then

k

=

lcm

(

m, n

)

6

=

mn

. Then

for any (

a, b

)

∈ C

n

× C

m

,we have (

a, b

)

k

= (

a

k

, b

k

) =

e

. So the order of any (

a, b

)

is at most

k < mn

. So there is no element of order

mn

. So

C

n

× C

m

is not a

cyclic group of order nm.

Given a complicated group

G

, it is sometimes helpful to write it as a product

H × K

, which could make things a bit simpler. We can do so by the following

theorem:

Proposition

(Direct product theorem)

.

Let

H

1

, H

2

≤ G

. Suppose the following

are true:

(i) H

1

∩ H

2

= {e}.

(ii) (∀a

i

∈ H

i

) a

1

a

2

= a

2

a

1

.

(iii) (∀a ∈ G)(∃a

i

∈ H

i

) a = a

1

a

2

. We also write this as G = H

1

H

2

.

Then G

∼

=

H

1

× H

2

.

Proof.

Define

f

:

H

1

×H

2

→ G

by

f

(

a

1

, a

2

) =

a

1

a

2

. Then it is a homomorphism

since

f((a

1

, a

2

) ∗ (b

1

, b

2

)) = f(a

1

b

1

, a

2

b

2

)

= a

1

b

1

a

2

b

2

= a

1

a

2

b

1

b

2

= f(a

1

, a

2

)f(b

1

, b

2

).

Surjectivity follows from (iii). We’ll show injectivity by showing that the kernel

is

{e}

. If

f

(

a

1

, a

2

) =

e

, then we know that

a

1

a

2

=

e

. Then

a

1

=

a

−1

2

. Since

a

1

∈ H

1

and

a

−1

2

∈ H

2

, we have

a

1

=

a

−1

2

∈ H

1

∩ H

2

=

{e}

. Thus

a

1

=

a

2

=

e

and ker f = {e}.