Third Isomorphism Theorem

After comparing

H/(H\cap K)

with

HK/K

, we now study what happens when one quotient is taken after another. The focus keyword third isomorphism theorem describes the rule

\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2, \end{equation*}

where

H_1\subseteq H_2

are normal subgroups of

G

. The theorem says that first collapsing

H_1

and then collapsing

H_2/H_1

has the same effect as collapsing

H_2

in one step. Students often find quotient-of-quotient notation intimidating, but the idea is only staged simplification.

Let

H_1\trianglelefteq G

and

H_2\trianglelefteq G

with

H_1\subseteq H_2

. Then

H_2/H_1\trianglelefteq G/H_1

, and

\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2. \end{equation*}

After comparing

H/(H\cap K)

with

HK/K

, we now study what happens when one quotient is taken after another. The focus keyword third isomorphism theorem describes the rule

\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2, \end{equation*}

where

H_1\subseteq H_2

are normal subgroups of

G

. The theorem says that first collapsing

H_1

and then collapsing

H_2/H_1

has the same effect as collapsing

H_2

in one step. Students often find quotient-of-quotient notation intimidating, but the idea is only staged simplification.

Let

H_1\trianglelefteq G

and

H_2\trianglelefteq G

with

H_1\subseteq H_2

. Then

H_2/H_1\trianglelefteq G/H_1

, and

\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2. \end{equation*}

Given that

H_1\trianglelefteq G

H_2\trianglelefteq G

, and

H_1\subseteq H_2

. To prove that

H_2/H_1\trianglelefteq G/H_1

and

(G/H_1)/(H_2/H_1)\cong G/H_2

. Define

\phi:G/H_1\to G/H_2

\begin{equation*} \phi(aH_1)=aH_2. \end{equation*}

[1] To prove that

\phi

is well-defined. Suppose

aH_1=bH_1

. Then

b^{-1}a\in H_1

. Since

H_1\subseteq H_2

, we have

b^{-1}a\in H_2

. Therefore

aH_2=bH_2

. Thus

\phi(aH_1)=\phi(bH_1)

. [2] To prove that

\phi

is a homomorphism. Let

aH_1,bH_1\in G/H_1

. Then

\begin{align*} \phi((aH_1)(bH_1))&=\phi(abH_1)\\ &=abH_2,\\ \phi(aH_1)\phi(bH_1)&=(aH_2)(bH_2)\\ &=abH_2. \end{align*}

Thus

\phi((aH_1)(bH_1))=\phi(aH_1)\phi(bH_1)

. [3] To prove that

\phi

is onto. Every element of

G/H_2

has the form

aH_2

, and

\begin{equation*} aH_2=\phi(aH_1). \end{equation*}

[4] To compute the kernel.

\begin{align*} \ker\phi&=\{aH_1\in G/H_1\mid \phi(aH_1)=H_2\}\\ &=\{aH_1\in G/H_1\mid aH_2=H_2\}\\ &=\{aH_1\mid a\in H_2\}\\ &=H_2/H_1. \end{align*}

Since kernels are normal subgroups,

H_2/H_1\trianglelefteq G/H_1

. By the first isomorphism theorem,

\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2. \end{equation*}

\square

The quotient

G/H_1

has already made every element of

H_1

into the identity coset. Inside this quotient, the subgroup

H_2/H_1

represents what remains of

H_2

. Collapsing that remaining subgroup finishes the job of collapsing all of

H_2

. Thus the third isomorphism theorem tells us that quotienting can be performed in stages without changing the final structure. This is especially useful when large quotient groups are easier to understand through smaller intermediate quotients.

(\mathbb{Z},+)

, let

H_1=(6)

and

H_2=(3)

. Since

(6)\subseteq(3)

and every subgroup of

\mathbb{Z}

is normal, the third isomorphism theorem gives

\begin{equation*} (\mathbb{Z}/(6))/((3)/(6))\cong\mathbb{Z}/(3). \end{equation*}

Thus a quotient of

\mathbb{Z}/(6)

by the subgroup

(3)/(6)

is isomorphic to

\mathbb{Z}_3

Let

N\subseteq H\subseteq G

with

N\trianglelefteq G

and

H\trianglelefteq G

. Then the third isomorphism theorem gives

\begin{equation*} (G/N)/(H/N)\cong G/H. \end{equation*}

The first quotient collapses

N

. The second quotient collapses the image of

H

inside

G/N

. Together, these two steps collapse exactly

H

in the original group.

This preview uses subgroups

(m)\subseteq(n)

\mathbb Z

, so the divisibility condition is

n\mid m

. First

\mathbb Z

is collapsed modulo

m

, and then the subgroup

(n)/(m)

is collapsed inside

\mathbb Z/(m)

. Adjust

m

and

n

and observe the size of the intermediate quotient, the subgroup being collapsed, and the final quotient. Try

m=12,n=4

and then enter a pair where

n

does not divide

m

to see why containment is required.

The middle group remembers residues modulo

m

. The subgroup

(n)/(m)

is what remains of the larger subgroup after the first collapse. Quotienting by that remaining part gives the same final structure as reducing modulo

n

from the start.

This calculator verifies the integer version of the third isomorphism theorem. Enter

m

and

n

with

n\mid m

, corresponding to

(m)\subseteq(n)

. The calculator reports the size of the intermediate group, the size of the subgroup being collapsed, and the size of the final quotient. Use it to confirm that quotienting by stages reduces the size from

m

n

exactly as

\mathbb Z/(n)

does.

The number of cosets after the second quotient is exactly the order of the larger quotient

\mathbb Z/(n)

. The theorem strengthens this size match into an isomorphism of groups.

Let

(m)\subseteq(n)\subseteq\mathbb{Z}

. Use the third isomorphism theorem to write a quotient of

\mathbb{Z}/(m)

isomorphic to

\mathbb{Z}/(n)

Let

(m)\subseteq(n)\subseteq\mathbb{Z}

. Then

n\mid m

. Take

H_1=(m)

and

H_2=(n)

. Since

\mathbb{Z}

is abelian, both subgroups are normal in

\mathbb{Z}

. By the third isomorphism theorem,

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

Thus the quotient of

\mathbb{Z}/(m)

by the subgroup

(n)/(m)

is isomorphic to

\mathbb{Z}/(n)

\square

Let

G

be a group and let

H_1\subseteq H_2

be normal subgroups of

G

. Identify the natural epimorphism from

G/H_1

onto

G/H_2

and its kernel.

Let

H_1\subseteq H_2

be normal subgroups of

G

. Define

\phi:G/H_1\to G/H_2

\begin{equation*} \phi(aH_1)=aH_2. \end{equation*}

This is well-defined because

aH_1=bH_1

implies

b^{-1}a\in H_1\subseteq H_2

, so

aH_2=bH_2

. The map is onto because every coset

aH_2

is the image of

aH_1

. Its kernel is

\begin{align*} \ker\phi&=\{aH_1\mid aH_2=H_2\}\\ &=\{aH_1\mid a\in H_2\}\\ &=H_2/H_1. \end{align*}

\square

In the third isomorphism theorem, check the containment direction carefully. The smaller normal subgroup

H_1

must be inside the larger normal subgroup

H_2

, so that

H_2/H_1

is a subgroup of

G/H_1

1. Apply the theorem to

(12)\subseteq(4)\subseteq\mathbb{Z}

. 2. If

H_1=H_2

, what does the theorem say? 3. If

H_1=\{e\}

, what does the theorem reduce to?

1. The theorem gives

(\mathbb{Z}/(12))/((4)/(12))\cong\mathbb{Z}/(4)

. 2. It gives

(G/H_1)/(H_1/H_1)\cong G/H_1

, quotienting by the trivial subgroup. 3. It gives

(G/\{e\})/(H_2/\{e\})\cong G/H_2

, which is essentially

G/H_2

It is the kernel of the natural epimorphism

G/H_1\to G/H_2

Quotienting by stages gives the same result as quotienting directly by the larger normal subgroup.

No. The quotient groups involved require normality.

BMLabs Mathematics

BMLabs Mathematics

Third Isomorphism Theorem

Third Isomorphism Theorem

QUOTIENTING IN STAGES

THIRD ISOMORPHISM THEOREM

WHY THE THEOREM IS NATURAL

QUOTIENTS OF THE INTEGERS

QUOTIENTING BY STAGES

EXPLORE QUOTIENTING IN STAGES

QUOTIENT STAGE EXPLORER

INTERPRET THE STAGES

COMPUTE THE THIRD ISOMORPHISM SIZES

QUOTIENT STAGE CALCULATOR

CONNECT THE CALCULATION

FREQUENTLY ASKED QUESTIONS

CONTINUE TO THE CORRESPONDENCE THEOREM

Coordinates and Locus

Invariants under Orthogonal Transformation