Homomorphic Images and Normal Subgroups

The first isomorphism theorem shows that every epimorphism from

G

produces a quotient of

G

. The focus keyword homomorphic images and normal subgroups captures this exact connection: homomorphic images of a group are the same, up to isomorphism, as quotient groups by normal subgroups. If

f:G\to G_1

is onto, then

G_1\cong G/\ker f

, and the kernel is normal in

G

. Conversely, every normal subgroup gives a natural homomorphism onto a quotient group. This lesson turns the first isomorphism theorem into a classification principle for possible epimorphic images.

Let

G

and

G_1

be groups. The group

G_1

is called a \textbf{homomorphic image} of

G

if there exists an epimorphism

f:G\to G_1

The first isomorphism theorem shows that every epimorphism from

G

produces a quotient of

G

. The focus keyword homomorphic images and normal subgroups captures this exact connection: homomorphic images of a group are the same, up to isomorphism, as quotient groups by normal subgroups. If

f:G\to G_1

is onto, then

G_1\cong G/\ker f

, and the kernel is normal in

G

. Conversely, every normal subgroup gives a natural homomorphism onto a quotient group. This lesson turns the first isomorphism theorem into a classification principle for possible epimorphic images.

Let

G

and

G_1

be groups. The group

G_1

is called a \textbf{homomorphic image} of

G

if there exists an epimorphism

f:G\to G_1

A group

G_1

is a homomorphic image of

G

if and only if there exists a normal subgroup

N\trianglelefteq G

such that

\begin{equation*} G/N\cong G_1. \end{equation*}

Given that

G

and

G_1

are groups. To prove that

G_1

is a homomorphic image of

G

if and only if

G/N\cong G_1

for some

N\trianglelefteq G

. First suppose that

G_1

is a homomorphic image of

G

. Then there exists an epimorphism

f:G\to G_1

. Put

N=\ker f

. Since kernels are normal subgroups,

N\trianglelefteq G

. By the first isomorphism theorem,

\begin{equation*} G/N=G/\ker f\cong G_1. \end{equation*}

Conversely, suppose there exists

N\trianglelefteq G

such that

G/N\cong G_1

. Let

\pi:G\to G/N

be the natural homomorphism. Then

\pi

is onto. Let

\alpha:G/N\to G_1

be an isomorphism. The composition

\begin{equation*} \alpha\circ\pi:G\to G_1 \end{equation*}

is an epimorphism, because both

\pi

and

\alpha

are onto. Therefore

G_1

is a homomorphic image of

G

. Hence the characterization is proved.

\square

A homomorphic image may be smaller or simpler than the original group, but some structure survives. If the original group is cyclic, the image is cyclic. If the original group is commutative, the image is commutative. These statements follow from onto-ness: every element in the image group comes from an element of the original group. A common mistake is to reverse these implications. A noncyclic group can have a cyclic quotient, and a noncommutative group can have a commutative quotient.

Let

f:G\to G_1

be an epimorphism. If

G

is cyclic, then

G_1

is cyclic.

Given that

f:G\to G_1

is an epimorphism and

G

is cyclic. To prove that

G_1

is cyclic. Let

G=\langle a\rangle

. Since

f

is onto, for every

y\in G_1

there exists

x\in G

such that

f(x)=y

. Since

G=\langle a\rangle

, there exists

n\in\mathbb{Z}

such that

x=a^n

. Hence

\begin{align*} y&=f(x)\\ &=f(a^n)\\ &=f(a)^n. \end{align*}

Thus every element of

G_1

is a power of

f(a)

. Therefore

G_1=\langle f(a)\rangle

. Hence

G_1

is cyclic.

\square

Let

f:G\to G_1

be an epimorphism. If

G

is commutative, then

G_1

is commutative.

Given that

f:G\to G_1

is an epimorphism and

G

is commutative. To prove that

G_1

is commutative. 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)\\ &=f(ba) \quad [\because G \text{ is commutative}]\\ &=f(b)f(a)\\ &=yx. \end{align*}

Hence

G_1

is commutative.

\square

The homomorphic images of

S_3

correspond to quotient groups

S_3/N

, where

N\trianglelefteq S_3

. The normal subgroups of

S_3

are

\begin{equation*} \{e\},\quad A_3,\quad S_3. \end{equation*}

Thus the homomorphic images, up to isomorphism, are

\begin{equation*} S_3/\{e\}\cong S_3,\qquad S_3/A_3\cong\mathbb{Z}_2,\qquad S_3/S_3\cong\mathbb{Z}_1. \end{equation*}

So the only homomorphic images of

S_3

, up to isomorphism, are

S_3

\mathbb{Z}_2

, and the trivial group.

This preview shows how a normal subgroup determines a homomorphic image. Select one of the normal subgroups of

S_3

and observe the quotient that it produces. The subgroup selected as the kernel is the part of

S_3

collapsed to the identity in the image. Try moving from

\{e\}

A_3

S_3

and watch the image become smaller. This is the classification principle: different normal kernels give the possible epimorphic images.

The selected normal subgroup plays the role of a kernel. Enlarging the kernel collapses more of the original group, so the resulting homomorphic image becomes smaller. This is why listing normal subgroups is the correct first step when listing homomorphic images.

This calculator tests the most common finite cyclic case. For cyclic groups, an epimorphism from

\mathbb Z_n

onto

\mathbb Z_m

exists exactly when

m

divides

n

. Enter

n

and

m

and compare the target size with the kernel size. The output explains why a cyclic source can only map onto cyclic targets whose order divides the source order.

The divisibility condition is the finite cyclic version of the normal-subgroup classification. A successful epimorphism has a kernel of the right size, and the quotient by that kernel has exactly the target order.

Show that there is no epimorphism from

\mathbb{Z}_4

onto the Klein four-group

V

Let

V

be the Klein four-group. If there were an epimorphism

f:\mathbb{Z}_4\to V

, then

V

would be a homomorphic image of

\mathbb{Z}_4

. Since

\mathbb{Z}_4

is cyclic, every homomorphic image of

\mathbb{Z}_4

must be cyclic. But

V

is not cyclic. Each nonidentity element of

V

has order

2

, so no element generates all four elements of

V

. This contradiction shows that no epimorphism from

\mathbb{Z}_4

onto

V

exists.

\square

Show that there is no epimorphism from the Klein four-group

V

onto

\mathbb{Z}_4

Let

V

be the Klein four-group. Suppose there exists an epimorphism

f:V\to\mathbb{Z}_4

. The group

\mathbb{Z}_4

has an element of order

4

, namely

[1]

. Since

f

is onto, there exists

x\in V

such that

f(x)=[1]

. Then

o(f(x))=4

. For every homomorphism, the order of the image of an element divides the order of the element. Hence

4\mid o(x)

. This is impossible because every element of

V

has order

1

2

. Therefore no epimorphism from

V

onto

\mathbb{Z}_4

exists.

\square

To list homomorphic images of a group, list its normal subgroups first. To rule out an epimorphism, compare structural features that must survive under epimorphic images, such as cyclicity, commutativity, and possible element orders.

1. List the homomorphic images of a simple group

G

, up to isomorphism. 2. Prove that every homomorphic image of an abelian group is abelian. 3. Can a nonabelian group have an abelian homomorphic image?

1. A simple group has only

\{e\}

and

G

as normal subgroups, so its homomorphic images are isomorphic to

G

and to the trivial group. 2. This is exactly the homomorphic image of a commutative group theorem. 3. Yes. For example,

S_3/A_3\cong\mathbb{Z}_2

Up to isomorphism, yes. Every homomorphic image is a quotient by a normal subgroup.

Kernels are normal, and quotient groups require normal subgroups.

It cannot gain properties that epimorphic images cannot support, such as cyclicity from a noncyclic source in reverse; structural implications must be checked carefully.

BMLabs Mathematics

BMLabs Mathematics

Homomorphic Images and Normal Subgroups

Homomorphic Images and Normal Subgroups

IMAGES CONTROLLED BY NORMAL SUBGROUPS

HOMOMORPHIC IMAGE

CHARACTERIZATION OF HOMOMORPHIC IMAGES

PRESERVED STRUCTURE UNDER EPIMORPHISMS

HOMOMORPHIC IMAGE OF A CYCLIC GROUP

HOMOMORPHIC IMAGE OF A COMMUTATIVE GROUP

HOMOMORPHIC IMAGES OF S THREE

EXPLORE QUOTIENTS FROM NORMAL SUBGROUPS

NORMAL KERNEL IMAGE EXPLORER

INTERPRET THE QUOTIENT IMAGE

COMPUTE CYCLIC EPIMORPHIC IMAGES

CYCLIC EPIMORPHISM CHECKER

CONNECT THE CALCULATION

FREQUENTLY ASKED QUESTIONS

CONTINUE TO THE SECOND ISOMORPHISM THEOREM

Coordinates and Locus

Invariants under Orthogonal Transformation