Product Formula for Finite Subgroups

The product of two subgroups is another important place where Lagrange's theorem and coset counting appear. If

H

and

K

are finite subgroups of a group

G

, the set

HK

consists of all products

h\circ k

with

h\in H

and

k\in K

. Even when

HK

is not itself a subgroup, its number of elements can still be counted exactly. The overlap between

H

and

K

controls the repeated representations of elements of

HK

. The result is the product formula

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

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. The

\textbf{product}

H

and

K

is the subset

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

The product of two subgroups is another important place where Lagrange's theorem and coset counting appear. If

H

and

K

are finite subgroups of a group

G

, the set

HK

consists of all products

h\circ k

with

h\in H

and

k\in K

. Even when

HK

is not itself a subgroup, its number of elements can still be counted exactly. The overlap between

H

and

K

controls the repeated representations of elements of

HK

. The result is the product formula

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

Let

(G,\circ)

be a group and let

H

and

K

be subgroups of

G

. The

\textbf{product}

H

and

K

is the subset

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

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,K

are finite subgroups of

G

. To prove that

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

Consider the map

\varphi:H\times K\to HK

defined by

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

By the definition of

HK

, the map

\varphi

is onto. We determine the number of preimages of a fixed element of

HK

. Let

x\in HK

. Then there exist

h\in H

and

k\in K

such that

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

Let

(h',k')\in H\times K

such that

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

Then

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

Therefore

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

Since

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

and

k\circ (k')^{-1}\in K

, we get

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

Let

\begin{equation*} t=h^{-1}\circ h'. \end{equation*}

Then

t\in H\cap K

and

\begin{equation*} h'=h\circ t. \end{equation*}

Also, from

h'\circ k'=h\circ k

, we get

\begin{align*} h\circ t\circ k'=h\circ k &\implies t\circ k'=k\\ &\implies k'=t^{-1}\circ k. \end{align*}

Thus each preimage of

x

has the form

\begin{equation*} (h\circ t,\;t^{-1}\circ k) \end{equation*}

where

t\in H\cap K

. Conversely, for every

t\in H\cap K

, the pair

\begin{equation*} (h\circ t,\;t^{-1}\circ k) \end{equation*}

belongs to

H\times K

, and

\begin{align*} \varphi(h\circ t,\;t^{-1}\circ k) &=(h\circ t)\circ(t^{-1}\circ k)\\ &=h\circ (t\circ t^{-1})\circ k\\ &=h\circ e\circ k\\ &=h\circ k\\ &=x. \end{align*}

Therefore every element of

HK

has exactly

|H\cap K|

preimages under

\varphi

. Since

\begin{equation*} |H\times K|=|H||K|, \end{equation*}

we get

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

Therefore

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

Hence,

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

\square

We observe the map from

H\times K

onto the product set

HK

. Choose the sizes of

H

K

, and

H\cap K

, and notice how the grid of ordered pairs is grouped into equal-size fibers. As

|H\cap K|

increases, more pairs represent the same product, so the number of distinct elements in

HK

decreases. Try setting

|H|=4

|K|=3

, and

|H\cap K|=1

to match the example in

\mathbb{Z}_{12}

The denominator

|H\cap K|

is not an arbitrary correction factor. It is the exact number of repeated representations of each product element, so dividing by it converts ordered pairs into distinct elements of

HK

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. If

H\cap K=\{e\}

, then

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

Given that

(G,\circ)

is a group,

H,K

are finite subgroups of

G

, and

H\cap K=\{e\}

. To prove that

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

Since

H\cap K=\{e\}

, we get

\begin{equation*} |H\cap K|=1. \end{equation*}

By the product formula for finite subgroups,

\begin{align*} |HK| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{|H||K|}{1}\\ &=|H||K|. \end{align*}

Hence,

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

\square

The formula counts the set

HK

. It does not automatically say that

HK

is a subgroup. In many courses, students mistakenly treat every product of subgroups as a subgroup. That is not valid without extra hypotheses. The counting formula remains true because it counts representations

h\circ k

, not because

HK

must be closed under the group operation.

Let

G=\mathbb{Z}_{12}

under addition modulo

12

. Let

\begin{equation*} H=\{0,3,6,9\} \end{equation*}

and

\begin{equation*} K=\{0,4,8\}. \end{equation*}

Then

\begin{equation*} H\cap K=\{0\}. \end{equation*}

Therefore

\begin{align*} |H+K| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{4\cdot 3}{1}\\ &=12. \end{align*}

Indeed,

\begin{equation*} H+K=\mathbb{Z}_{12}. \end{equation*}

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. Suppose

\begin{equation*} |H|=12,\quad |K|=10,\quad |H\cap K|=2. \end{equation*}

Find

|HK|

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. By the product formula for finite subgroups,

\begin{align*} |HK| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{12\cdot 10}{2}\\ &=60. \end{align*}

Therefore

\begin{equation*} |HK|=60. \end{equation*}

We compute

|HK|

from the three numbers that control the product formula. Enter the orders of

H

K

, and their intersection, and check whether the division produces an integer. The output reminds us that the formula counts the subset

HK

, not necessarily a subgroup. This distinction keeps the calculation connected to the proof rather than to an unproved closure assumption.

1. Let

H,K

be finite subgroups of a group

G

such that

|H|=8

|K|=6

, and

|H\cap K|=2

. Find

|HK|

. 2. Let

H,K

be finite subgroups of a group

G

such that

H\cap K=\{e\}

|H|=5

, and

|K|=7

. Find

|HK|

. 3. Let

H,K

be finite subgroups of a group

G

. Prove that

|H\cap K|

divides

|H||K|

1. By the product formula,

\begin{equation*} |HK|=\frac{8\cdot 6}{2}=24. \end{equation*}

2. Since

H\cap K=\{e\}

, we get

\begin{equation*} |HK|=|H||K|=5\cdot 7=35. \end{equation*}

3. Since

H\cap K

is a finite subgroup,

|H\cap K|

is a positive integer. The product formula gives

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

Therefore

|H\cap K|

divides

|H||K|

No. The product

HK

is always a subset, but it need not be a subgroup without extra conditions.

Elements in

H\cap K

account for repeated representations of the same element of

HK

It requires

H

and

K

to be finite. The whole group

G

need not be finite for this counting formula.

BMLabs Mathematics

BMLabs Mathematics

Product Formula for Finite Subgroups

Product Formula for Finite Subgroups

OVERVIEW

PRODUCT OF TWO SUBGROUPS

EXPLORE REPEATED REPRESENTATIONS

SUBGROUP PRODUCT REPRESENTATION EXPLORER

WHAT THE FIBERS EXPLAIN

PRODUCT FORMULA IN INTEGERS MODULO TWELVE

COUNTING A PRODUCT OF SUBGROUPS

SOLUTION

CALCULATE THE PRODUCT SIZE

PRODUCT FORMULA CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation