Products of Subgroups

After intersections and unions, the next natural construction is the product of two subgroups. The focus keyword for this lecture is product of subgroups. If

H

and

K

are subgroups of a group

G

, then

HK

is the set of all products

h\circ k

with

h\in H

and

k\in K

. Unlike intersections, the product

HK

is not always a subgroup. In this lesson, we prove the exact condition

HK=KH

and derive the counting formula for finite subgroups.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. The \textbf{product of subgroups}

H

and

K

is the subset

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

Similarly,

\begin{equation*} KH=\{k\circ h:k\in K,\;h\in H\}. \end{equation*}

After intersections and unions, the next natural construction is the product of two subgroups. The focus keyword for this lecture is product of subgroups. If

H

and

K

are subgroups of a group

G

, then

HK

is the set of all products

h\circ k

with

h\in H

and

k\in K

. Unlike intersections, the product

HK

is not always a subgroup. In this lesson, we prove the exact condition

HK=KH

and derive the counting formula for finite subgroups.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. The \textbf{product of subgroups}

H

and

K

is the subset

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

Similarly,

\begin{equation*} KH=\{k\circ h:k\in K,\;h\in H\}. \end{equation*}

In an abelian group,

h\circ k=k\circ h

, so

HK=KH

automatically. In a non-abelian group, the order of factors matters, and

HK

may differ from

KH

. This is why the subgroup property of

HK

requires a special condition. Students often assume that a product of subgroups must be a subgroup because the notation resembles multiplication. The theorem below shows exactly when this assumption is valid.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. Then

HK

is a subgroup of

G

if and only if

\begin{equation*} HK=KH. \end{equation*}

Given that

(G,\circ)

is a group and

H

and

K

are two subgroups of

G

. Let

\begin{equation*} HK=\{h\circ k:h\in H,\;k\in K\} \end{equation*}

and

\begin{equation*} KH=\{k\circ h:k\in K,\;h\in H\}. \end{equation*}

To prove that

HK

is a subgroup of

G

if and only if

HK=KH

. [1] Let

HK

be a subgroup of

G

. To prove that

HK=KH

. Let

x\in HK

. Then there exist

h\in H

and

k\in K

such that

\begin{equation*} x=h\circ k. \end{equation*}

Since

HK

is a subgroup of

G

, we get

x^{-1}\in HK

. Thus there exist

h_1\in H

and

k_1\in K

such that

\begin{equation*} x^{-1}=h_1\circ k_1. \end{equation*}

Taking inverse in

G

, we get

\begin{align*} x &=(h_1\circ k_1)^{-1}\\ &=k_1^{-1}\circ h_1^{-1}. \end{align*}

Since

k_1^{-1}\in K

and

h_1^{-1}\in H

, we get

x\in KH

. Therefore

HK\subseteq KH

. Let

y\in KH

. Then there exist

k\in K

and

h\in H

such that

\begin{equation*} y=k\circ h. \end{equation*}

Now

\begin{align*} y^{-1} &=h^{-1}\circ k^{-1}\in HK. \end{align*}

Since

HK

is a subgroup of

G

, we get

y=(y^{-1})^{-1}\in HK

. Therefore

KH\subseteq HK

. Hence,

HK=KH

. [2] Let

HK=KH

. To prove that

HK

is a subgroup of

G

. Since

e\in H

and

e\in K

, we get

\begin{equation*} e=e\circ e\in HK. \end{equation*}

Thus

HK

is non-empty. Let

x,y\in HK

. Then there exist

h_1,h_2\in H

and

k_1,k_2\in K

such that

\begin{equation*} x=h_1\circ k_1,\quad y=h_2\circ k_2. \end{equation*}

Now

\begin{align*} x\circ y^{-1} &=(h_1\circ k_1)\circ(h_2\circ k_2)^{-1}\\ &=h_1\circ k_1\circ k_2^{-1}\circ h_2^{-1}. \end{align*}

Since

k_1\circ k_2^{-1}\in K

and

h_2^{-1}\in H

, we get

\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}\in KH. \end{equation*}

Using

HK=KH

, we get

\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}\in HK. \end{equation*}

Thus there exist

h_3\in H

and

k_3\in K

such that

\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}=h_3\circ k_3. \end{equation*}

Therefore

\begin{align*} x\circ y^{-1} &=h_1\circ h_3\circ k_3. \end{align*}

Since

h_1\circ h_3\in H

and

k_3\in K

, we get

\begin{equation*} x\circ y^{-1}\in HK. \end{equation*}

By the one-step subgroup criterion,

HK

is a subgroup of

G

. Hence,

HK

is a subgroup of

G

if and only if

HK=KH

\square

Let

(G,\circ)

be an abelian group and let

H

and

K

be two subgroups of

G

. Then

HK

is a subgroup of

G

Given that

(G,\circ)

is an abelian group and

H

and

K

are two subgroups of

G

. To prove that

HK

is a subgroup of

G

. Let

x\in HK

. Then there exist

h\in H

and

k\in K

such that

\begin{equation*} x=h\circ k. \end{equation*}

Since

G

is abelian,

\begin{align*} x &=h\circ k\\ &=k\circ h. \end{align*}

Therefore

x\in KH

, and hence

HK\subseteq KH

. Similarly,

KH\subseteq HK

. Thus

HK=KH

. By the product theorem,

HK

is a subgroup of

G

. Hence,

HK\leq G

\square

Let

(G,\circ)

be a group and let

H

and

K

be finite subgroups of

G

. Then

\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}

Given that

(G,\circ)

is a group and

H

and

K

are finite subgroups of

G

. To prove that

\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}

Define

f:H\times K\to HK

\begin{equation*} f(h,k)=h\circ k. \end{equation*}

Then

f

is onto by definition of

HK

. Let

h\in H

and

k\in K

. Suppose

\begin{equation*} h_1\circ k_1=h\circ k. \end{equation*}

Then

\begin{align*} h_1\circ k_1=h\circ k &\implies h^{-1}\circ h_1=k\circ k_1^{-1}. \end{align*}

Since

h^{-1}\circ h_1\in H

and

k\circ k_1^{-1}\in K

, we get

\begin{equation*} h^{-1}\circ h_1\in H\cap K. \end{equation*}

Let

\begin{equation*} t=h^{-1}\circ h_1. \end{equation*}

Then

t\in H\cap K

and

\begin{equation*} h_1=h\circ t,\quad k_1=t^{-1}\circ k. \end{equation*}

Conversely, for each

t\in H\cap K

\begin{align*} (h\circ t)\circ(t^{-1}\circ k) &=h\circ k. \end{align*}

Therefore each element of

HK

has exactly

|H\cap K|

inverse images under

f

. Thus

\begin{align*} |H||K| &=|HK||H\cap K|\\ &\implies |HK|=\frac{|H||K|}{|H\cap K|}. \end{align*}

Hence,

\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}

\square

We observe

HK

and

KH

inside the symmetric group

S_3

. Choose two small subgroups, and the preview lists the two product sets. If the sets are equal, the theorem says

HK

is a subgroup. If the sets differ, the product set is not a subgroup, even though both

H

and

K

are subgroups.

The product set is not just a union and not just a pairwise list. In non-abelian groups, reversing the order can change the product set and therefore change whether the product is a subgroup.

Let

(G,\circ)

be a group such that

\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}

Prove that

H=\{x^6:x\in G\}

is a subgroup of

G

Let

(G,\circ)

be a group such that

\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}

Let

\begin{equation*} H=\{x^6:x\in G\}. \end{equation*}

To prove that

H

is a subgroup of

G

. Since

e=e^6

, we get

e\in H

. Therefore

H

is non-empty. From the earlier result, the set

\begin{equation*} K=\{x^2:x\in G\} \end{equation*}

is a subgroup of

G

. Let

x^6,y^6\in H

, where

x,y\in G

. Then

x^2,y^2\in K

. Using the given condition with

a=x^2

and

b=y^2

, we get

\begin{align*} (x^2\circ y^2)^3 &=(x^2)^3\circ (y^2)^3\\ &=x^6\circ y^6. \end{align*}

Since

K

is a subgroup of

G

and

x^2,y^2\in K

, we get

\begin{equation*} x^2\circ y^2\in K. \end{equation*}

Therefore there exists

z\in G

such that

\begin{equation*} x^2\circ y^2=z^2. \end{equation*}

Thus

\begin{align*} x^6\circ y^6 &=(x^2\circ y^2)^3\\ &=(z^2)^3\\ &=z^6\in H. \end{align*}

Therefore

H

is closed under

\circ

. Let

x^6\in H

. Then

\begin{align*} (x^6)^{-1} &=x^{-6}\\ &=(x^{-1})^6\in H. \end{align*}

Therefore

(x^6)^{-1}\in H

for every

x^6\in H

. By the subgroup criterion,

H

is a subgroup of

G

. Hence,

H\leq G

\square

For finite subgroups, the size of

HK

depends on the overlap

H\cap K

. Enter the orders of

H

K

, and their intersection. The calculator checks divisibility and then applies

|HK|=|H||K|/|H\cap K|

. This formula counts the product set even when

HK

is not itself a subgroup.

1. Let

G

be abelian and let

H,K\leq G

. Prove directly that

HK\leq G

. 2. Let

H,K

be finite subgroups of

G

with

|H|=6

|K|=10

, and

|H\cap K|=2

. Find

|HK|

. 3. If

H\cap K=\{e\}

and

H,K

are finite subgroups, prove that

|HK|=|H||K|

1. Since

G

is abelian,

HK=KH

. Therefore

HK

is a subgroup by the product theorem. 2.

|HK|=\frac{6\cdot 10}{2}=30

. 3. Since

|H\cap K|=1

, the cardinality formula gives

|HK|=\frac{|H||K|}{1}=|H||K|

No. It is a subgroup exactly when

HK=KH

Yes. Since all elements commute, the two products are equal.

No. The formula counts the product set

HK

when

H

and

K

are finite subgroups.

BMLabs Mathematics

BMLabs Mathematics

Products of Subgroups

Products of Subgroups

PRODUCTS OF SUBGROUPS

PRODUCT OF SUBGROUPS

ORDER OF THE FACTORS

PRODUCT OF TWO SUBGROUPS

PROOF OF PRODUCT OF TWO SUBGROUPS

PRODUCT IN AN ABELIAN GROUP

PROOF OF PRODUCT IN AN ABELIAN GROUP

CARDINALITY OF A PRODUCT

PROOF OF CARDINALITY OF A PRODUCT

EXPLORE PRODUCTS IN A NON-ABELIAN GROUP

SUBGROUP PRODUCT EXPLORER

ORDER MATTERS IN PRODUCTS

SIXTH POWERS UNDER A CUBIC CONDITION

SOLUTION OF SIXTH POWERS UNDER A CUBIC CONDITION

COMPUTE THE SIZE OF A PRODUCT SET

PRODUCT CARDINALITY CALCULATOR

EXERCISES ON PRODUCTS OF SUBGROUPS

ANSWERS ON PRODUCTS OF SUBGROUPS

FREQUENTLY ASKED QUESTIONS

PROCEED TO SPECIAL SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation