Isomorphism Tests and Examples

After studying epimorphisms, monomorphisms, and isomorphisms, we now need reliable tests for deciding when two groups are structurally the same. The focus keyword isomorphism tests and examples refers to two complementary tasks. To prove that two groups are isomorphic, we construct a bijective homomorphism. To prove that two groups are not isomorphic, we compare properties that every isomorphism must preserve. Cardinality, commutativity, cyclicity, and orders of elements are among the most useful invariants. Students often try to prove non-isomorphism by saying that the elements look different, but appearance is irrelevant. What matters is whether the group structure can be matched exactly.

A proof of isomorphism has three parts: homomorphism, one-to-one, and onto. The kernel test often shortens the one-to-one part, because a homomorphism is one-to-one exactly when its kernel is identity-only. A proof of non-isomorphism should point to a property preserved by isomorphism. If one group is commutative and the other is not, they cannot be isomorphic. If one group has an element of order

4

and the other has no element of order

4

, they cannot be isomorphic. The strongest isomorphism tests and examples make the preserved property explicit.

4

and the other has no element of order

4

, they cannot be isomorphic. The strongest isomorphism tests and examples make the preserved property explicit.

Let

f:G\to G_1

be an isomorphism. Then

f^{-1}:G_1\to G

is also an isomorphism.

Given that

f:G\to G_1

is an isomorphism. To prove that

f^{-1}:G_1\to G

is an isomorphism. Since

f

is one-to-one and onto, the inverse function

f^{-1}

exists and is also one-to-one and onto. It remains to prove that

f^{-1}

is a homomorphism. Let

x,y\in G_1

. Since

f

is onto, there exist

a,b\in G

such that

x=f(a)

and

y=f(b)

. Then

\begin{align*} xy&=f(a)f(b)\\ &=f(ab). \end{align*}

Applying

f^{-1}

gives

\begin{align*} f^{-1}(xy)&=ab\\ &=f^{-1}(x)f^{-1}(y). \end{align*}

Thus

f^{-1}

is a homomorphism. Since it is one-to-one and onto,

f^{-1}

is an isomorphism.

\square

Let

f:G\to G_1

be an isomorphism. Then commutativity, cyclicity, cardinality, and orders of elements are preserved by

f

Given that

f:G\to G_1

is an isomorphism. To prove that commutativity, cyclicity, cardinality, and element orders are preserved. Since

f

is bijective,

G

and

G_1

have the same cardinality. Suppose

G

is commutative. Let

x,y\in G_1

. Since

f

is onto,

x=f(a)

and

y=f(b)

for some

a,b\in G

. Then

\begin{align*} xy&=f(a)f(b)\\ &=f(ab)\\ &=f(ba)\\ &=f(b)f(a)\\ &=yx. \end{align*}

Thus

G_1

is commutative. The reverse direction follows by applying the same argument to

f^{-1}

. Let

a\in G

and suppose

o(a)=n<\infty

. Then

a^n=e

, so

f(a)^n=e_1

. Thus

o(f(a))\mid n

. Applying the same argument to

f^{-1}

gives

n\mid o(f(a))

. Therefore

o(f(a))=o(a)

. If

a

has infinite order and

f(a)

had finite order

r

, then

f(a)^r=e_1

, so

f(a^r)=e_1

. Since

f

is one-to-one,

a^r=e

, a contradiction. Hence infinite order is also preserved. If

G=\langle a\rangle

, then every element of

G_1

has the form

f(a^m)=f(a)^m

. Hence

G_1=\langle f(a)\rangle

. The converse follows from

f^{-1}

. Hence the listed properties are preserved by isomorphism.

\square

The group

(\mathbb{Z},+)

is cyclic because

\mathbb{Z}=\langle 1\rangle

. The group

(\mathbb{Q},+)

is not cyclic. If

\mathbb{Q}=\langle q\rangle

for some

q\in\mathbb{Q}

, then every rational number would be an integral multiple of

q

. But

q/2\in\mathbb{Q}

, and

kq=q/2

would force

k=1/2

, not an integer. Since cyclicity is preserved by isomorphism,

\begin{equation*} (\mathbb{Z},+)\not\cong(\mathbb{Q},+). \end{equation*}

The group

(\mathbb{Q},+)

has no nonzero element of finite order. Indeed, if

nq=0

with

n>0

, then

q=0

. In

(\mathbb{Q}^{\ast},\cdot)

, the element

-1

has order

2

because

(-1)^2=1

and

-1\neq 1

. Since isomorphisms preserve orders of elements,

\begin{equation*} (\mathbb{Q},+)\not\cong(\mathbb{Q}^{\ast},\cdot). \end{equation*}

Show that

(\mathbb{R}^{\ast},\cdot)

and

(\mathbb{C}^{\ast},\cdot)

are not isomorphic.

Let

\mathbb{R}^{\ast}=\mathbb{R}\setminus\{0\}

and

\mathbb{C}^{\ast}=\mathbb{C}\setminus\{0\}

under multiplication. In

\mathbb{C}^{\ast}

, the element

i

satisfies

\begin{equation*} i^4=1 \end{equation*}

and no smaller positive power of

i

equals

1

. Hence

o(i)=4

. In

\mathbb{R}^{\ast}

, suppose

x^4=1

. Then

\begin{equation*} (x^2-1)(x^2+1)=0. \end{equation*}

Since

x\in\mathbb{R}^{\ast}

, this gives

x=1

x=-1

. The element

1

has order

1

, and the element

-1

has order

2

. Thus

\mathbb{R}^{\ast}

has no element of order

4

. Since element orders are preserved by isomorphism, the two groups are not isomorphic. Hence

(\mathbb{R}^{\ast},\cdot)\not\cong(\mathbb{C}^{\ast},\cdot)

\square

Let

G

be a group and define

f:G\to G

f(a)=a^{-1}

. Prove that

f

is a homomorphism if and only if

G

is commutative.

Let

G

be a group, and let

f:G\to G

be defined by

f(a)=a^{-1}

. To prove that

f

is a homomorphism if and only if

G

is commutative. First suppose that

f

is a homomorphism. Let

a,b\in G

. Then

\begin{align*} (ab)^{-1}&=f(ab)\\ &=f(a)f(b)\\ &=a^{-1}b^{-1}. \end{align*}

But in every group,

(ab)^{-1}=b^{-1}a^{-1}

. Therefore

\begin{equation*} b^{-1}a^{-1}=a^{-1}b^{-1}. \end{equation*}

Taking inverses of both sides gives

ab=ba

. Hence

G

is commutative. Conversely, suppose

G

is commutative. Then for

a,b\in G

\begin{align*} f(ab)&=(ab)^{-1}\\ &=b^{-1}a^{-1}\\ &=a^{-1}b^{-1}\\ &=f(a)f(b). \end{align*}

Hence

f

is a homomorphism.

\square

When disproving isomorphism, do not compare notation. Compare invariants. The most efficient questions are: do the groups have the same size, are both commutative, are both cyclic, and do they have the same numbers of elements of each possible order?

1. Prove that

\mathbb{Z}_4

and

\mathbb{Z}_2\times\mathbb{Z}_2

are not isomorphic. 2. Show that any group isomorphic to a cyclic group is cyclic. 3. If

G\cong H

and

G

has exactly three elements of order

2

, what can be said about

H

1. The group

\mathbb{Z}_4

has an element of order

4

, but every nonidentity element of

\mathbb{Z}_2\times\mathbb{Z}_2

has order

2

. 2. If

G=\langle a\rangle

and

f:G\to H

is an isomorphism, then

H=\langle f(a)\rangle

. 3. The group

H

also has exactly three elements of order

2

, because isomorphisms preserve element orders and biject elements.

Yes. For example,

\mathbb{Z}_4

and

\mathbb{Z}_2\times\mathbb{Z}_2

both have order

4

but are not isomorphic.

The image of a commutative group is commutative. An isomorphism preserves commutativity in both directions.

An isomorphism sends each element to an element of the same order, so a mismatch in element orders disproves isomorphism.

BMLabs Mathematics

BMLabs Mathematics

Isomorphism Tests and Examples

Isomorphism Tests and Examples

PROVING AND DISPROVING STRUCTURAL SAMENESS

THE STANDARD ISOMORPHISM STRATEGY

INVERSE OF AN ISOMORPHISM

PROPERTIES PRESERVED BY ISOMORPHISM

INTEGERS AND RATIONALS ARE NOT ISOMORPHIC

ADDITIVE AND MULTIPLICATIVE RATIONAL GROUPS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO CYCLIC CLASSIFICATION

Coordinates and Locus

Invariants under Orthogonal Transformation