AATA Definition and Examples

Section 9.1 Definition and Examples

Two groups

(G, \cdot)

and

(H, \circ)

are isomorphic if there exists a one-to-one and onto map

ϕ : G \to H

such that the group operation is preserved; that is,

ϕ (a \cdot b) = ϕ (a) \circ ϕ (b)

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for all

a

and

b

G .

G

is isomorphic to

H,

we write

G ≅ H .

The map

ϕ

is called an isomorphism.

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Example 9.1.

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To show that

Z_{4} ≅ ⟨ i ⟩,

define a map

ϕ : Z_{4} \to ⟨ i ⟩

ϕ (n) = i^{n} .

We must show that

ϕ

is bijective and preserves the group operation. The map

ϕ

is one-to-one and onto because

\begin{aligned} ϕ (0) & = 1 \\ ϕ (1) & = i \\ ϕ (2) & = - 1 \\ ϕ (3) & = - i . \end{aligned}

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Since

ϕ (m + n) = i^{m + n} = i^{m} i^{n} = ϕ (m) ϕ (n),

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the group operation is preserved.

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Example 9.2.

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We can define an isomorphism

ϕ

from the additive group of real numbers

(R, +)

to the multiplicative group of positive real numbers

(R^{+}, \cdot)

with the exponential map; that is,

ϕ (x + y) = e^{x + y} = e^{x} e^{y} = ϕ (x) ϕ (y) .

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Of course, we must still show that

ϕ

is one-to-one and onto, but this can be determined using calculus.

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Example 9.3.

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The integers are isomorphic to the subgroup of

Q^{*}

consisting of elements of the form

2^{n} .

Define a map

ϕ : Z \to Q^{*}

ϕ (n) = 2^{n} .

Then

ϕ (m + n) = 2^{m + n} = 2^{m} 2^{n} = ϕ (m) ϕ (n) .

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By definition the map

ϕ

is onto the subset

{2^{n} : n \in Z}

Q^{*} .

To show that the map is injective, assume that

m \neq n .

If we can show that

ϕ (m) \neq ϕ (n),

then we are done. Suppose that

m > n

and assume that

ϕ (m) = ϕ (n) .

Then

2^{m} = 2^{n}

2^{m - n} = 1,

which is impossible since

m - n > 0 .

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Example 9.4.

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The groups

Z_{8}

and

Z_{12}

cannot be isomorphic since they have different orders; however, it is true that

U (8) ≅ U (12) .

We know that

\begin{aligned} U (8) & = {1, 3, 5, 7} \\ U (12) & = {1, 5, 7, 11} . \end{aligned}

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An isomorphism

ϕ : U (8) \to U (12)

is then given by

\begin{aligned} 1 & \mapsto 1 \\ 3 & \mapsto 5 \\ 5 & \mapsto 7 \\ 7 & \mapsto 11 . \end{aligned}

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The map

ϕ

is not the only possible isomorphism between these two groups. We could define another isomorphism

ψ

ψ (1) = 1,

ψ (3) = 11,

ψ (5) = 5,

ψ (7) = 7 .

In fact, both of these groups are isomorphic to

Z_{2} \times Z_{2}

(see Example 3.28 in Chapter 3).

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Example 9.5.

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Even though

S_{3}

and

Z_{6}

possess the same number of elements, we would suspect that they are not isomorphic, because

Z_{6}

is abelian and

S_{3}

is nonabelian. To demonstrate that this is indeed the case, suppose that

ϕ : Z_{6} \to S_{3}

is an isomorphism. Let

a, b \in S_{3}

be two elements such that

a b \neq b a .

Since

ϕ

is an isomorphism, there exist elements

m

and

n

Z_{6}

such that

ϕ (m) = a and ϕ (n) = b .

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However,

a b = ϕ (m) ϕ (n) = ϕ (m + n) = ϕ (n + m) = ϕ (n) ϕ (m) = b a,

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which contradicts the fact that

a

and

b

do not commute.

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Theorem 9.6.

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Let

ϕ : G \to H

be an isomorphism of two groups. Then the following statements are true.

$ϕ^{- 1} : H \to G$ is an isomorphism.
$| G | = | H | .$
If $G$ is abelian, then $H$ is abelian.
If $G$ is cyclic, then $H$ is cyclic.
If $G$ has a subgroup of order $n,$ then $H$ has a subgroup of order $n .$

Proof.

Assertions (1) and (2) follow from the fact that

ϕ

is a bijection. We will prove (3) here and leave the remainder of the theorem to be proved in the exercises.

(3) Suppose that

h_{1}

and

h_{2}

are elements of

H .

Since

ϕ

is onto, there exist elements

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

such that

ϕ (g_{1}) = h_{1}

and

ϕ (g_{2}) = h_{2} .

Therefore,

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

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We are now in a position to characterize all cyclic groups.

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Theorem 9.7.

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All cyclic groups of infinite order are isomorphic to

Z .

Proof.

Let

G

be a cyclic group with infinite order and suppose that

a

is a generator of

G .

Define a map

ϕ : Z \to G

ϕ : n \mapsto a^{n} .

Then

ϕ (m + n) = a^{m + n} = a^{m} a^{n} = ϕ (m) ϕ (n) .

To show that

ϕ

is injective, suppose that

m

and

n

are two elements in

Z,

where

m \neq n .

We can assume that

m > n .

We must show that

a^{m} \neq a^{n} .

Let us suppose the contrary; that is,

a^{m} = a^{n} .

In this case

a^{m - n} = e,

where

m - n > 0,

which contradicts the fact that

a

has infinite order. Our map is onto since any element in

G

can be written as

a^{n}

for some integer

n

and

ϕ (n) = a^{n} .

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Theorem 9.8.

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G

is a cyclic group of order

n,

then

G

is isomorphic to

Z_{n} .

Proof.

Let

G

be a cyclic group of order

n

generated by

a

and define a map

ϕ : Z_{n} \to G

ϕ : k \mapsto a^{k},

where

0 \leq k < n .

The proof that

ϕ

is an isomorphism is one of the end-of-chapter exercises.

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Corollary 9.9.

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G

is a group of order

p,

where

p

is a prime number, then

G

is isomorphic to

Z_{p} .

Proof.

The proof is a direct result of Corollary 6.12.

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The main goal in group theory is to classify all groups; however, it makes sense to consider two groups to be the same if they are isomorphic. We state this result in the following theorem, whose proof is left as an exercise.

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Theorem 9.10.

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The isomorphism of groups determines an equivalence relation on the class of all groups.

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Hence, we can modify our goal of classifying all groups to classifying all groups up to isomorphism; that is, we will consider two groups to be the same if they are isomorphic.

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Subsection Cayley’s Theorem

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Cayley proved that if

G

is a group, it is isomorphic to a group of permutations on some set; hence, every group is a permutation group. Cayley’s Theorem is what we call a representation theorem. The aim of representation theory is to find an isomorphism of some group

G

that we wish to study into a group that we know a great deal about, such as a group of permutations or matrices.

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Example 9.11.

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Consider the group

Z_{3} .

The Cayley table for

Z_{3}

is as follows.

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\begin{array}{cccc} + & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}

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The addition table of

Z_{3}

suggests that it is the same as the permutation group

G = {(0), (012), (021)} .

The isomorphism here is

\begin{aligned} 0 & \mapsto (\begin{array}{c} 0 & 1 & 2 \\ 0 & 1 & 2 \end{array}) = (0) \\ 1 & \mapsto (\begin{array}{c} 0 & 1 & 2 \\ 1 & 2 & 0 \end{array}) = (0 1 2) \\ 2 & \mapsto (\begin{array}{c} 0 & 1 & 2 \\ 2 & 0 & 1 \end{array}) = (0 2 1) . \end{aligned}

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Theorem 9.12. Cayley.

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Every group is isomorphic to a group of permutations.

Proof.

Let

G

be a group. We must find a group of permutations

\overset{―}{G}

that is isomorphic to

G .

For any

g \in G,

define a function

λ_{g} : G \to G

λ_{g} (a) = g a .

We claim that

λ_{g}

is a permutation of

G .

To show that

λ_{g}

is one-to-one, suppose that

λ_{g} (a) = λ_{g} (b) .

Then

g a = λ_{g} (a) = λ_{g} (b) = g b .

Hence,

a = b .

To show that

λ_{g}

is onto, we must prove that for each

a \in G,

there is a

b

such that

λ_{g} (b) = a .

Let

b = g^{- 1} a .

Now we are ready to define our group

\overset{―}{G} .

Let

\overset{―}{G} = {λ_{g} : g \in G} .

We must show that

\overset{―}{G}

is a group under composition of functions and find an isomorphism between

G

and

\overset{―}{G} .

We have closure under composition of functions since

(λ_{g} \circ λ_{h}) (a) = λ_{g} (h a) = g h a = λ_{g h} (a) .

Also,

λ_{e} (a) = e a = a

and

(λ_{g^{- 1}} \circ λ_{g}) (a) = λ_{g^{- 1}} (g a) = g^{- 1} g a = a = λ_{e} (a) .

We can define an isomorphism from

G

\overset{―}{G}

ϕ : g \mapsto λ_{g} .

The group operation is preserved since

ϕ (g h) = λ_{g h} = λ_{g} λ_{h} = ϕ (g) ϕ (h) .

It is also one-to-one, because if

ϕ (g) (a) = ϕ (h) (a),

then

g a = λ_{g} a = λ_{h} a = h a .

Hence,

g = h .

That

ϕ

is onto follows from the fact that

ϕ (g) = λ_{g}

for any

λ_{g} \in \overset{―}{G} .

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The isomorphism

g \mapsto λ_{g}

is known as the left regular representation of

G .

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Subsection Historical Note

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Arthur Cayley was born in England in 1821, though he spent much of the first part of his life in Russia, where his father was a merchant. Cayley was educated at Cambridge, where he took the first Smith’s Prize in mathematics. A lawyer for much of his adult life, he wrote several papers in his early twenties before entering the legal profession at the age of 25. While practicing law he continued his mathematical research, writing more than 300 papers during this period of his life. These included some of his best work. In 1863 he left law to become a professor at Cambridge. Cayley wrote more than 900 papers in fields such as group theory, geometry, and linear algebra. His legal knowledge was very valuable to Cambridge; he participated in the writing of many of the university’s statutes. Cayley was also one of the people responsible for the admission of women to Cambridge.

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