First Isomorphism Theorem

Having established factorization through quotients, we now choose the most important possible quotient: the quotient by the kernel. The focus keyword first isomorphism theorem says that

G/\ker f

has exactly the same structure as the image

f(G)

. The kernel is precisely the part of

G

collapsed to the identity, so quotienting by it removes exactly the ambiguity created by

f

. Students often memorize the formula but miss the idea: the quotient identifies exactly those elements that have the same image. This theorem is the central bridge from homomorphisms to quotient groups.

Let

f:G\to G_1

be a homomorphism. Then

\begin{equation*} G/\ker f\cong f(G). \end{equation*}

f

is onto, then

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

Having established factorization through quotients, we now choose the most important possible quotient: the quotient by the kernel. The focus keyword first isomorphism theorem says that

G/\ker f

has exactly the same structure as the image

f(G)

. The kernel is precisely the part of

G

collapsed to the identity, so quotienting by it removes exactly the ambiguity created by

f

Let

f:G\to G_1

be a homomorphism. Then

\begin{equation*} G/\ker f\cong f(G). \end{equation*}

f

is onto, then

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

Given that

f:G\to G_1

is a homomorphism. To prove that

G/\ker f\cong f(G)

, and that

G/\ker f\cong G_1

when

f

is onto. Let

N=\ker f

. Since kernels are normal subgroups,

N\trianglelefteq G

. Define

\Phi:G/N\to f(G)

\begin{equation*} \Phi(aN)=f(a). \end{equation*}

[1] To prove that

\Phi

is well-defined. Suppose

aN=bN

. Then

b^{-1}a\in N=\ker f

. Therefore

f(b^{-1}a)=e_1

, and

\begin{align*} f(b)^{-1}f(a)&=e_1,\\ f(a)&=f(b). \end{align*}

Thus

\Phi(aN)=\Phi(bN)

. [2] To prove that

\Phi

is a homomorphism. Let

aN,bN\in G/N

. Then

\begin{align*} \Phi((aN)(bN))&=\Phi(abN)\\ &=f(ab)\\ &=f(a)f(b)\\ &=\Phi(aN)\Phi(bN). \end{align*}

[3] To prove that

\Phi

is onto. Let

y\in f(G)

. Then

y=f(a)

for some

a\in G

. Hence

y=\Phi(aN)

. [4] To prove that

\Phi

is one-to-one. If

\Phi(aN)=e_1

, then

f(a)=e_1

, so

a\in\ker f=N

. Therefore

aN=N

. Thus

\ker\Phi=\{N\}

, and

\Phi

is one-to-one. Since

\Phi

is a one-to-one and onto homomorphism, it is an isomorphism. Hence

\begin{equation*} G/\ker f\cong f(G). \end{equation*}

f

is onto, then

f(G)=G_1

, so

G/\ker f\cong G_1

\square

The first isomorphism theorem measures a homomorphism in two parts: what it kills and what it reaches. The kernel tells us which elements become identity after applying

f

, while the image tells us which elements of the codomain are actually reached. Quotienting by the kernel repairs the loss of injectivity. The result is not always the whole codomain; it is the image. Only when

f

is onto may we replace

f(G)

G_1

Let

f:(\mathbb{Z},+)\to(\mathbb{Z}_3,+_3)

be defined by

f(n)=[n]_3

. The kernel is

\begin{equation*} \ker f=(3). \end{equation*}

The map is onto because every element of

\mathbb{Z}_3

[n]_3

for some

n\in\mathbb{Z}

. By the first isomorphism theorem,

\begin{equation*} \mathbb{Z}/(3)\cong\mathbb{Z}_3. \end{equation*}

This shows that the usual residue class group is exactly the quotient of

\mathbb{Z}

by the subgroup of multiples of

3

Let

G=\langle a\rangle

be a cyclic group and define

f:\mathbb{Z}\to G

f(n)=a^n

. Then

\begin{equation*} f(m+n)=a^{m+n}=a^ma^n=f(m)f(n), \end{equation*}

f

is a homomorphism. It is onto because

G=\langle a\rangle

. If

o(a)=n

, then

\ker f=(n)

, and the first isomorphism theorem gives

\begin{equation*} \mathbb{Z}/(n)\cong G. \end{equation*}

a

has infinite order, then

\ker f=\{0\}

and

\mathbb{Z}\cong G

This preview models a homomorphism

\phi:\mathbb{Z}_n\to\mathbb{Z}_m

by the rule

\phi([x])=[kx]_m

. Choose

n

m

, and the multiplier

k

, then observe how the elements of

\mathbb{Z}_n

group into fibers with the same image. The theorem says that each fiber is a coset of the kernel, so collapsing the kernel leaves exactly one representative for each image value. Start with

n=12

m=8

, and

k=2

, then try changing

k

to see how the image and kernel change together.

Each row is one image element and the domain elements that map to it. The row containing

[0]

is the kernel, and every other row is a coset of that kernel. The first isomorphism theorem says that these rows, not the original elements, are what survive in the quotient.

This calculator checks the same finite cyclic model and verifies the counting statement behind the theorem. Enter

n

m

, and

k

for the rule

[x]_n\mapsto[kx]_m

. The rule is accepted only when it is well-defined as a homomorphism. The output compares the kernel size, image size, and quotient size so you can see why

G/\ker f

has the same structure size as

f(G)

The equality of sizes is not the theorem by itself, but it reveals the mechanism. The kernel identifies exactly the elements that should be treated as equivalent, and the quotient has one class for each image value.

Let

f:G\to G_1

be an onto homomorphism. Prove that

G/\ker f\cong G_1

Let

f:G\to G_1

be an onto homomorphism. By the first isomorphism theorem,

\begin{equation*} G/\ker f\cong f(G). \end{equation*}

Since

f

is onto, every element of

G_1

lies in

f(G)

. Hence

\begin{equation*} f(G)=G_1. \end{equation*}

Substituting this into the isomorphism gives

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

\square

Let

f:S_3\to\{1,-1\}

be the sign homomorphism. Use the first isomorphism theorem to identify

S_3/\ker f

Let

f:S_3\to\{1,-1\}

be the sign homomorphism. The kernel of

f

consists of the even permutations, so

\begin{equation*} \ker f=A_3. \end{equation*}

The map is onto because both

1

and

-1

occur as signs of permutations in

S_3

. By the first isomorphism theorem,

\begin{equation*} S_3/A_3\cong\{1,-1\}. \end{equation*}

Since

\{1,-1\}

under multiplication is cyclic of order

2

, we also have

\begin{equation*} S_3/A_3\cong\mathbb{Z}_2. \end{equation*}

\square

When applying the first isomorphism theorem, compute the kernel and the image separately. Do not assume the image is the full codomain unless the homomorphism is known to be onto.

1. Let

f:\mathbb{Z}\to\mathbb{Z}_6

be given by

f(n)=[n]_6

. Use the first isomorphism theorem. 2. Let

f:\mathbb{R}\to\mathbb{R}^{+}

f(x)=e^x

. Identify

\ker f

and the quotient isomorphic to the image. 3. If

f:G\to G_1

is one-to-one, what does the first isomorphism theorem say?

1. Since

\ker f=(6)

and

f

is onto,

\mathbb{Z}/(6)\cong\mathbb{Z}_6

. 2. The kernel is

\{0\}

and the image is

\mathbb{R}^{+}

, so

\mathbb{R}/\{0\}\cong\mathbb{R}^{+}

. 3. If

f

is one-to-one, then

\ker f=\{e\}

, so

G/\{e\}\cong f(G)

, hence

G\cong f(G)

The kernel is exactly the part of the domain collapsed to the identity, so quotienting by it removes precisely that collapse.

It is about the image. The codomain appears only when the homomorphism is onto.

Yes. Every homomorphic image is isomorphic to a quotient by the kernel.

BMLabs Mathematics

BMLabs Mathematics

First Isomorphism Theorem

First Isomorphism Theorem

QUOTIENT BY THE KERNEL

FIRST ISOMORPHISM THEOREM

WHAT THE THEOREM MEASURES

THE NATURAL MAP FROM Z TO Z THREE

CYCLIC GROUPS AS QUOTIENTS OF Z

EXPLORE FIBERS AND QUOTIENT CLASSES

KERNEL FIBER EXPLORER

INTERPRET THE FIBERS

COMPUTE KERNEL IMAGE BALANCE

FIRST ISOMORPHISM CALCULATOR

CONNECT THE CALCULATION

FREQUENTLY ASKED QUESTIONS

CONTINUE TO HOMOMORPHIC IMAGES

Introduction To Discrete Mathematics

Paths and Circuits