Second Isomorphism Theorem

After using normal subgroups to describe homomorphic images, we now compare a subgroup with a quotient involving a normal subgroup. The focus keyword second isomorphism theorem refers to the isomorphism

H/(H\cap K)\cong HK/K

, where

H\leq G

and

K\trianglelefteq G

. The theorem says that when

H

is combined with

K

, the part of

H

already lying inside

K

is exactly the part that disappears in the quotient. Students often find the notation dense, but the idea is simple: first restrict to

H

, then collapse the overlap with

K

Let

H

and

K

be subgroups of a group

G

. The product set

HK

is defined by

\begin{equation*} HK=\{hk\mid h\in H,\;k\in K\}. \end{equation*}

K\trianglelefteq G

, then

HK

is a subgroup of

G

H/(H\cap K)\cong HK/K

, where

H\leq G

and

K\trianglelefteq G

. The theorem says that when

H

is combined with

K

, the part of

H

already lying inside

K

is exactly the part that disappears in the quotient. Students often find the notation dense, but the idea is simple: first restrict to

H

, then collapse the overlap with

K

Let

H

and

K

be subgroups of a group

G

. The product set

HK

is defined by

\begin{equation*} HK=\{hk\mid h\in H,\;k\in K\}. \end{equation*}

K\trianglelefteq G

, then

HK

is a subgroup of

G

Let

H\leq G

and let

K\trianglelefteq G

. Then

H\cap K\trianglelefteq H

K\trianglelefteq HK

, and

\begin{equation*} H/(H\cap K)\cong HK/K. \end{equation*}

Given that

H\leq G

and

K\trianglelefteq G

. To prove that

H\cap K\trianglelefteq H

K\trianglelefteq HK

, and

H/(H\cap K)\cong HK/K

. Since

K\trianglelefteq G

, the product

HK

is a subgroup of

G

, and

K\trianglelefteq HK

. First prove

H\cap K\trianglelefteq H

. Let

x\in H\cap K

and

h\in H

. Since

x,h\in H

, we have

hxh^{-1}\in H

. Since

x\in K

and

K\trianglelefteq G

, we also have

hxh^{-1}\in K

. Therefore

hxh^{-1}\in H\cap K

. Hence

H\cap K\trianglelefteq H

. Define

\phi:H\to HK/K

\begin{equation*} \phi(h)=hK. \end{equation*}

[1] To prove that

\phi

is a homomorphism. For

h_1,h_2\in H

\begin{align*} \phi(h_1h_2)&=h_1h_2K,\\ \phi(h_1)\phi(h_2)&=(h_1K)(h_2K)\\ &=h_1h_2K. \end{align*}

Thus

\phi(h_1h_2)=\phi(h_1)\phi(h_2)

. [2] To compute the kernel.

\begin{align*} \ker\phi&=\{h\in H\mid hK=K\}\\ &=\{h\in H\mid h\in K\}\\ &=H\cap K. \end{align*}

[3] To compute the image. Every element of

HK/K

has the form

hkK

with

h\in H

and

k\in K

. Since

kK=K

, we have

hkK=hK=\phi(h)

. Therefore

\phi(H)=HK/K

. By the first isomorphism theorem,

\begin{equation*} H/\ker\phi\cong\phi(H). \end{equation*}

Substituting

\ker\phi=H\cap K

and

\phi(H)=HK/K

gives

\begin{equation*} H/(H\cap K)\cong HK/K. \end{equation*}

\square

The quotient

HK/K

records the part added to

K

by elements of

H

. However, any element of

H

already lying in

K

becomes invisible after quotienting by

K

. That invisible part is precisely

H\cap K

. Therefore

H/(H\cap K)

and

HK/K

carry the same information. The second isomorphism theorem is often called the diamond theorem because the subgroups

H

K

H\cap K

, and

HK

form a useful diagram of inclusions.

Let

H=(2)

and

K=(3)

(\mathbb{Z},+)

. Since

\mathbb{Z}

is abelian, every subgroup is normal. Here

\begin{equation*} H+K=(2)+(3)=\mathbb{Z} \end{equation*}

and

\begin{equation*} H\cap K=(6). \end{equation*}

By the second isomorphism theorem,

\begin{equation*} (2)/(6)\cong\mathbb{Z}/(3). \end{equation*}

Thus

(2)/(6)

has the same group structure as

\mathbb{Z}_3

This preview specializes the second isomorphism theorem to subgroups

(m)

and

(n)

\mathbb Z

. The top of the diamond is the sum

(m)+(n)=(\gcd(m,n))

, and the bottom is the intersection

(m)\cap(n)=(\operatorname{lcm}(m,n))

. Adjust

m

and

n

and observe how the overlap changes. The two side quotients have the same number of cosets, which reflects the isomorphism. Try

m=4,n=6

and then

m=2,n=3

to compare a nontrivial overlap with the example above.

The lower node is what

H

and

K

share, and the upper node is what they generate together. The theorem compares moving from

H

down to the overlap with moving from

H+K

down to

K

. In the integer case, the matching quotient orders make the abstract isomorphism visible.

This calculator verifies the second isomorphism theorem for integer subgroups. Enter positive integers

m

and

n

. It computes

(m)+(n)

using the greatest common divisor and

(m)\cap(n)

using the least common multiple. The two quotient orders should match every time, because

(m)/(\operatorname{lcm}(m,n))

and

(\gcd(m,n))/(n)

describe the same quotient structure.

The numerator on the left is

H

, so only the portion of

K

inside

H

can disappear there. On the right, quotienting

H+K

by all of

K

removes the same information. That is why the intersection appears in one denominator and

K

appears in the other.

Let

H=(m)

and

K=(n)

be subgroups of

(\mathbb{Z},+)

. Describe

H+K

and

H\cap K

Let

H=(m)

and

K=(n)

be subgroups of

\mathbb{Z}

. The subgroup generated by both

m

and

n

is generated by their greatest common divisor. Hence

\begin{equation*} (m)+(n)=(\gcd(m,n)). \end{equation*}

The intersection consists of all integers divisible by both

m

and

n

. Hence it is generated by the least common multiple:

\begin{equation*} (m)\cap(n)=(\operatorname{lcm}(m,n)). \end{equation*}

Therefore the second isomorphism theorem gives

\begin{equation*} (m)/(\operatorname{lcm}(m,n))\cong (\gcd(m,n))/(n), \end{equation*}

where the quotient on the right is interpreted inside

(\gcd(m,n))

\square

Let

H\leq G

and

K\trianglelefteq G

. Prove directly that

H\cap K\trianglelefteq H

Let

x\in H\cap K

and let

h\in H

. Since

x\in H

and

h\in H

, the subgroup property of

H

gives

\begin{equation*} hxh^{-1}\in H. \end{equation*}

Since

x\in K

and

K\trianglelefteq G

, conjugation by any element of

G

, and therefore by

h

, keeps

x

inside

K

. Hence

\begin{equation*} hxh^{-1}\in K. \end{equation*}

Thus

hxh^{-1}\in H\cap K

. Hence

H\cap K\trianglelefteq H

\square

In the second isomorphism theorem, the kernel of the map

h\mapsto hK

is not all of

K

; it is the part of

K

that lies inside

H

. That is why the denominator is

H\cap K

1. In

\mathbb{Z}

, apply the second isomorphism theorem to

H=(4)

and

K=(6)

. 2. If

H\cap K=\{e\}

, what does the theorem say? 3. If

H\subseteq K

, what does the theorem reduce to?

H+K=(2)

and

H\cap K=(12)

, so

(4)/(12)\cong(2)/(6)

. 2. It gives

H\cong HK/K

. 3. If

H\subseteq K

, then

HK=K

and

H\cap K=H

, so both quotient groups are trivial.

Normality ensures

HK

is a subgroup and that

HK/K

is a quotient group.

No. The theorem only needs

H\leq G

and

K\trianglelefteq G

It is

\phi:H\to HK/K

defined by

\phi(h)=hK

BMLabs Mathematics

BMLabs Mathematics

Second Isomorphism Theorem

Second Isomorphism Theorem

COMPARING SUBGROUPS AND QUOTIENTS

PRODUCT OF SUBGROUPS

SECOND ISOMORPHISM THEOREM

HOW TO READ THE THEOREM

INTEGER SUBGROUPS

EXPLORE THE DIAMOND FOR INTEGER SUBGROUPS

INTEGER DIAMOND EXPLORER

INTERPRET THE DIAMOND

COMPUTE THE GCD LCM FORM

GCD LCM ISOMORPHISM CALCULATOR

CONNECT THE CALCULATION

FREQUENTLY ASKED QUESTIONS

CONTINUE TO THE THIRD ISOMORPHISM THEOREM

Introduction To Discrete Mathematics

Paths and Circuits