AATA Direct Products

Section 9.2 Direct Products

Given two groups

G

and

H,

it is possible to construct a new group from the Cartesian product of

G

and

H,

G \times H .

Conversely, given a large group, it is sometimes possible to decompose the group; that is, a group is sometimes isomorphic to the direct product of two smaller groups. Rather than studying a large group

G,

it is often easier to study the component groups of

G .

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Subsection External Direct Products

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(G, \cdot)

and

(H, \circ)

are groups, then we can make the Cartesian product of

G

and

H

into a new group. As a set, our group is just the ordered pairs

(g, h) \in G \times H

where

g \in G

and

h \in H .

We can define a binary operation on

G \times H

(g_{1}, h_{1}) (g_{2}, h_{2}) = (g_{1} \cdot g_{2}, h_{1} \circ h_{2});

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that is, we just multiply elements in the first coordinate as we do in

G

and elements in the second coordinate as we do in

H .

We have specified the particular operations

\cdot

and

\circ

in each group here for the sake of clarity; we usually just write

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

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Proposition 9.13.

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Let

G

and

H

be groups. The set

G \times H

is a group under the operation

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

where

g_{1}, g_{2} \in G

and

h_{1}, h_{2} \in H .

Proof.

Clearly the binary operation defined above is closed. If

e_{G}

and

e_{H}

are the identities of the groups

G

and

H

respectively, then

(e_{G}, e_{H})

is the identity of

G \times H .

The inverse of

(g, h) \in G \times H

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

The fact that the operation is associative follows directly from the associativity of

G

and

H .

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Example 9.14.

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Let

R

be the group of real numbers under addition. The Cartesian product of

R

with itself,

R \times R = R^{2},

is also a group, in which the group operation is just addition in each coordinate; that is,

(a, b) + (c, d) = (a + c, b + d) .

The identity is

(0, 0)

and the inverse of

(a, b)

(- a, - b) .

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Example 9.15.

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Consider

Z_{2} \times Z_{2} = {(0, 0), (0, 1), (1, 0), (1, 1)} .

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Although

Z_{2} \times Z_{2}

and

Z_{4}

both contain four elements, they are not isomorphic. Every element

(a, b)

Z_{2} \times Z_{2}

other than the identity has order

2,

since

(a, b) + (a, b) = (0, 0);

however,

Z_{4}

is cyclic.

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The group

G \times H

is called the external direct product of

G

and

H .

Notice that there is nothing special about the fact that we have used only two groups to build a new group. The direct product

\prod_{i = 1}^{n} G_{i} = G_{1} \times G_{2} \times \dots \times G_{n}

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of the groups

G_{1}, G_{2}, \dots, G_{n}

is defined in exactly the same manner. If

G = G_{1} = G_{2} = \dots = G_{n},

we often write

G^{n}

instead of

G_{1} \times G_{2} \times \dots \times G_{n} .

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Example 9.16.

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The group

Z_{2}^{n},

considered as a set, is just the set of all binary

n

-tuples. The group operation is the “exclusive or” of two binary

n

-tuples. For example,

(01011101) + (01001011) = (00010110) .

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This group is important in coding theory, in cryptography, and in many areas of computer science.

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Theorem 9.17.

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Let

(g, h) \in G \times H .

g

and

h

have finite orders

r

and

s

respectively, then the order of

(g, h)

G \times H

is the least common multiple of

r

and

s .

Proof.

Suppose that

m

is the least common multiple of

r

and

s

and let

n = | (g, h) | .

Then

\begin{matrix} (g, h)^{m} = (g^{m}, h^{m}) = (e_{G}, e_{H}) \\ (g^{n}, h^{n}) = (g, h)^{n} = (e_{G}, e_{H}) . \end{matrix}

Hence,

n

must divide

m,

and

n \leq m .

However, by the second equation, both

r

and

s

must divide

n;

therefore,

n

is a common multiple of

r

and

s .

Since

m

is the least common multiple of

r

and

s,

m \leq n .

Consequently,

m

must be equal to

n .

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Corollary 9.18.

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Let

(g_{1}, \dots, g_{n}) \in \prod G_{i} .

g_{i}

has finite order

r_{i}

G_{i},

then the order of

(g_{1}, \dots, g_{n})

\prod G_{i}

is the least common multiple of

r_{1}, \dots, r_{n} .

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Example 9.19.

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Let

(8, 56) \in Z_{12} \times Z_{60} .

Since

gcd (8, 12) = 4,

the order of

8

12 / 4 = 3

Z_{12} .

Similarly, the order of

56

Z_{60}

15 .

The least common multiple of

3

and

15

15;

hence,

(8, 56)

has order

15

Z_{12} \times Z_{60} .

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Example 9.20.

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The group

Z_{2} \times Z_{3}

consists of the pairs

\begin{aligned} (0, 0), & (0, 1), & (0, 2), & (1, 0), & (1, 1), & (1, 2) . \end{aligned}

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In this case, unlike that of

Z_{2} \times Z_{2}

and

Z_{4},

it is true that

Z_{2} \times Z_{3} ≅ Z_{6} .

We need only show that

Z_{2} \times Z_{3}

is cyclic. It is easy to see that

(1, 1)

is a generator for

Z_{2} \times Z_{3} .

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The next theorem tells us exactly when the direct product of two cyclic groups is cyclic.

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Theorem 9.21.

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The group

Z_{m} \times Z_{n}

is isomorphic to

Z_{m n}

if and only if

gcd (m, n) = 1 .

Proof.

We will first show that if

Z_{m} \times Z_{n} ≅ Z_{m n},

then

gcd (m, n) = 1 .

We will prove the contrapositive; that is, we will show that if

gcd (m, n) = d > 1,

then

Z_{m} \times Z_{n}

cannot be cyclic. Notice that

m n / d

is divisible by both

m

and

n;

hence, for any element

(a, b) \in Z_{m} \times Z_{n},

\underset{m n / d times}{\underset{⏟}{(a, b) + (a, b) + \dots + (a, b)}} = (0, 0) .

Therefore, no

(a, b)

can generate all of

Z_{m} \times Z_{n} .

The converse follows directly from Theorem 9.17 since

lcm (m, n) = m n

if and only if

gcd (m, n) = 1 .

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Corollary 9.22.

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Let

n_{1}, \dots, n_{k}

be positive integers. Then

\prod_{i = 1}^{k} Z_{n_{i}} ≅ Z_{n_{1} \dots n_{k}}

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if and only if

gcd (n_{i}, n_{j}) = 1

for

i \neq j .

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Corollary 9.23.

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m = p_{1}^{e_{1}} \dots p_{k}^{e_{k}},

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where the

p_{i}

s are distinct primes, then

Z_{m} ≅ Z_{p_{1}^{e_{1}}} \times \dots \times Z_{p_{k}^{e_{k}}} .

Proof.

Since the greatest common divisor of

p_{i}^{e_{i}}

and

p_{j}^{e_{j}}

is 1 for

i \neq j,

the proof follows from Corollary 9.22.

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In Chapter 13, we will prove that all finite abelian groups are isomorphic to direct products of the form

Z_{p_{1}^{e_{1}}} \times \dots \times Z_{p_{k}^{e_{k}}}

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where

p_{1}, \dots, p_{k}

are (not necessarily distinct) primes.

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Subsection Internal Direct Products

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The external direct product of two groups builds a large group out of two smaller groups. We would like to be able to reverse this process and conveniently break down a group into its direct product components; that is, we would like to be able to say when a group is isomorphic to the direct product of two of its subgroups.

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Let

G

be a group with subgroups

H

and

K

satisfying the following conditions.

$G = H K = {h k : h \in H, k \in K};$
$H \cap K = {e};$
$h k = k h$ for all $k \in K$ and $h \in H .$

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Then

G

is the internal direct product of

H

and

K .

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Example 9.24.

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The group

U (8)

is the internal direct product of

H = {1, 3} and K = {1, 5} .

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Example 9.25.

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The dihedral group

D_{6}

is an internal direct product of its two subgroups

H = {id, r^{3}} and K = {id, r^{2}, r^{4}, s, r^{2} s, r^{4} s} .

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It can easily be shown that

K ≅ S_{3};

consequently,

D_{6} ≅ Z_{2} \times S_{3} .

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Example 9.26.

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Not every group can be written as the internal direct product of two of its proper subgroups. If the group

S_{3}

were an internal direct product of its proper subgroups

H

and

K,

then one of the subgroups, say

H,

would have to have order

3 .

In this case

H

is the subgroup

{(1), (123), (132)} .

The subgroup

K

must have order

2,

but no matter which subgroup we choose for

K,

the condition that

h k = k h

will never be satisfied for

h \in H

and

k \in K .

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Theorem 9.27.

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Let

G

be the internal direct product of subgroups

H

and

K .

Then

G

is isomorphic to

H \times K .

Proof.

Since

G

is an internal direct product, we can write any element

g \in G

g = h k

for some

h \in H

and some

k \in K .

Define a map

ϕ : G \to H \times K

ϕ (g) = (h, k) .

The first problem that we must face is to show that

ϕ

is a well-defined map; that is, we must show that

h

and

k

are uniquely determined by

g .

Suppose that

g = h k = h^{'} k^{'} .

Then

h^{- 1} h^{'} = k (k^{'})^{- 1}

is in both

H

and

K,

so it must be the identity. Therefore,

h = h^{'}

and

k = k^{'},

which proves that

ϕ

is, indeed, well-defined.

To show that

ϕ

preserves the group operation, let

g_{1} = h_{1} k_{1}

and

g_{2} = h_{2} k_{2}

and observe that

\begin{aligned} ϕ (g_{1} g_{2}) & = ϕ (h_{1} k_{1} h_{2} k_{2}) \\ = ϕ (h_{1} h_{2} k_{1} k_{2}) \\ = (h_{1} h_{2}, k_{1} k_{2}) \\ = (h_{1}, k_{1}) (h_{2}, k_{2}) \\ = ϕ (g_{1}) ϕ (g_{2}) . \end{aligned}

We will leave the proof that

ϕ

is one-to-one and onto as an exercise.

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Example 9.28.

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The group

Z_{6}

is an internal direct product isomorphic to

{0, 2, 4} \times {0, 3} .

🔗

We can extend the definition of an internal direct product of

G

to a collection of subgroups

H_{1}, H_{2}, \dots, H_{n}

G,

by requiring that

$G = H_{1} H_{2} \dots H_{n} = {h_{1} h_{2} \dots h_{n} : h_{i} \in H_{i}};$
$H_{i} \cap ⟨ \cup_{j \neq i} H_{j} ⟩ = {e};$
$h_{i} h_{j} = h_{j} h_{i}$ for all $h_{i} \in H_{i}$ and $h_{j} \in H_{j} .$

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We will leave the proof of the following theorem as an exercise.

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Theorem 9.29.

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Let

G

be the internal direct product of subgroups

H_{i},

where

i = 1, 2, \dots, n .

Then

G

is isomorphic to

\prod_{i} H_{i} .

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