Cardinality of Double Cosets

After proving that double cosets are equivalence classes, we now count their elements. The ordinary coset formula says that every left or right coset of a subgroup has the same size as the subgroup. A double coset has two subgroups acting at once, so the count is more subtle. The repeated representations of an element

h\circ a\circ k

are controlled by the subgroup

H\cap aKa^{-1}

. In this lecture, we prove the double coset cardinality formula and use it to compute examples.

Let

(G,\circ)

be a group and let

K

be a subgroup of

G

. For

a\in G

, the

\textbf{conjugate subgroup}

K

a

is the subset

\begin{equation*} aKa^{-1}=\{a\circ k\circ a^{-1}:k\in K\}. \end{equation*}

h\circ a\circ k

are controlled by the subgroup

H\cap aKa^{-1}

. In this lecture, we prove the double coset cardinality formula and use it to compute examples.

Let

(G,\circ)

be a group and let

K

be a subgroup of

G

. For

a\in G

, the

\textbf{conjugate subgroup}

K

a

is the subset

\begin{equation*} aKa^{-1}=\{a\circ k\circ a^{-1}:k\in K\}. \end{equation*}

Let

(G,\circ)

be a group and let

K

be a subgroup of

G

. If

a\in G

, then

aKa^{-1}

is a subgroup of

G

Given that

(G,\circ)

is a group,

K

is a subgroup of

G

, and

a\in G

. To prove that

aKa^{-1}

is a subgroup of

G

. Since

e\in K

, we get

\begin{align*} a\circ e\circ a^{-1} &=a\circ a^{-1}\\ &=e. \end{align*}

Therefore

\begin{equation*} e\in aKa^{-1}. \end{equation*}

Let

x,y\in aKa^{-1}

. Then there exist

k_1,k_2\in K

such that

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

and

\begin{equation*} y=a\circ k_2\circ a^{-1}. \end{equation*}

Now

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

Since

K

is a subgroup of

G

and

k_1,k_2\in K

, we get

\begin{equation*} k_1\circ k_2^{-1}\in K. \end{equation*}

Therefore

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

By the subgroup test,

aKa^{-1}

is a subgroup of

G

. Hence,

aKa^{-1}

is a subgroup of

G

\square

Let

(G,\circ)

be a group and let

H

and

K

be finite subgroups of

G

. If

a\in G

, then

\begin{equation*} |HaK|=\frac{|H||K|}{|H\cap aKa^{-1}|}. \end{equation*}

Given that

(G,\circ)

is a group,

H

and

K

are finite subgroups of

G

, and

a\in G

. To prove that

\begin{equation*} |HaK|=\frac{|H||K|}{|H\cap aKa^{-1}|}. \end{equation*}

Define

f:H\times K\to HaK

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

for all

h\in H

and

k\in K

. By the definition of

HaK

, the map

f

is onto. Fix

h\in H

and

k\in K

. Suppose

\begin{equation*} f(h_1,k_1)=f(h,k). \end{equation*}

Then

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

Let

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

Then

t\in H

. Also

\begin{equation*} t=a\circ k\circ k_1^{-1}\circ a^{-1}. \end{equation*}

Since

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

, we get

\begin{equation*} t\in aKa^{-1}. \end{equation*}

Therefore

\begin{equation*} t\in H\cap aKa^{-1}. \end{equation*}

From

t=h^{-1}\circ h_1

, we get

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

Since

t\in aKa^{-1}

, there exists

u\in K

such that

\begin{equation*} t=a\circ u\circ a^{-1}. \end{equation*}

From

\begin{equation*} t=a\circ k\circ k_1^{-1}\circ a^{-1}, \end{equation*}

we get

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

Since

t=a\circ u\circ a^{-1}

, we have

\begin{equation*} u=a^{-1}\circ t\circ a. \end{equation*}

Therefore

\begin{equation*} k_1=(a^{-1}\circ t^{-1}\circ a)\circ k. \end{equation*}

Thus every preimage of

f(h,k)

has the form

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

for some

t\in H\cap aKa^{-1}

. Conversely, let

t\in H\cap aKa^{-1}

. Then

h\circ t\in H

. Since

t\in aKa^{-1}

, we get

\begin{equation*} a^{-1}\circ t^{-1}\circ a\in K. \end{equation*}

Therefore

\begin{equation*} (a^{-1}\circ t^{-1}\circ a)\circ k\in K. \end{equation*}

Now

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

Therefore every element of

HaK

has exactly

\begin{equation*} |H\cap aKa^{-1}| \end{equation*}

preimages under

f

. Since

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

we get

\begin{align*} |H||K| &=|HaK||H\cap aKa^{-1}|. \end{align*}

Therefore

\begin{equation*} |HaK|=\frac{|H||K|}{|H\cap aKa^{-1}|}. \end{equation*}

Hence,

\begin{equation*} |HaK|=\frac{|H||K|}{|H\cap aKa^{-1}|}. \end{equation*}

\square

The cardinality formula comes from counting how many pairs

(h,k)\in H\times K

produce the same element of

HaK

. In an additive cyclic group, the conjugate subgroup

aKa^{-1}

behaves like

K

, so the overlap is

H\cap K

. Enter finite subsets

H

and

K

\mathbb{Z}_n

, and the preview groups all pairs

(h,k)

by the value of

h+a+k

. Large groups of repeated outputs show why

|H||K|

must be divided by an intersection size. The displayed comparison connects the formula to the actual computed set.

The preview shows the map from

H\times K

HaK

by grouping pairs with the same output. The denominator measures the repetition in that map, so the final number of distinct outputs is smaller when many pairs represent the same element.

Let

(G,\circ)

be a group and let

H

and

K

be finite subgroups of

G

. If

\begin{equation*} H\cap aKa^{-1}=\{e\}, \end{equation*}

then

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

Given that

(G,\circ)

is a group,

H

and

K

are finite subgroups of

G

a\in G

, and

\begin{equation*} H\cap aKa^{-1}=\{e\}. \end{equation*}

To prove that

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

Since

H\cap aKa^{-1}=\{e\}

, we get

\begin{equation*} |H\cap aKa^{-1}|=1. \end{equation*}

By the double coset cardinality formula,

\begin{align*} |HaK| &=\frac{|H||K|}{|H\cap aKa^{-1}|}\\ &=\frac{|H||K|}{1}\\ &=|H||K|. \end{align*}

Hence,

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

\square

Let

G=\mathbb{Z}_{12}

under addition modulo

12

. Let

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

and

\begin{equation*} K=\{0,6\}. \end{equation*}

Since

G

is abelian, the conjugate subgroup corresponding to

K

is just

K

. Thus

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

For

a=1

\begin{align*} |H+1+K| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{3\cdot 2}{1}\\ &=6. \end{align*}

This agrees with the computed double coset

\begin{equation*} H+1+K=\{1,3,5,7,9,11\}. \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 aKa^{-1}|=2. \end{equation*}

Find

|HaK|

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. By the double coset cardinality formula,

\begin{align*} |HaK| &=\frac{|H||K|}{|H\cap aKa^{-1}|}\\ &=\frac{12\cdot 10}{2}\\ &=60. \end{align*}

Therefore

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

The double coset formula is often used after the subgroup sizes and intersection size are already known. Enter the three sizes into the calculator and compare the substitution with the final value. The denominator must be positive, and the quotient should be an integer in a valid finite group situation. If the quotient is not an integer, the entered data cannot describe a valid double coset count. This calculator is useful for checking numerical examples before writing the final solution.

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. Suppose

\begin{equation*} H\cap aKa^{-1}=\{e\}, \end{equation*}

\begin{equation*} |H|=7, \end{equation*}

and

\begin{equation*} |K|=5. \end{equation*}

Find

|HaK|

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. Since

\begin{equation*} H\cap aKa^{-1}=\{e\}, \end{equation*}

we get

\begin{equation*} |H\cap aKa^{-1}|=1. \end{equation*}

By the double coset cardinality formula,

\begin{align*} |HaK| &=\frac{|H||K|}{|H\cap aKa^{-1}|}\\ &=\frac{7\cdot 5}{1}\\ &=35. \end{align*}

Therefore

\begin{equation*} |HaK|=35. \end{equation*}

1. Let

H,K

be finite subgroups of a group

G

. Suppose

|H|=8

|K|=6

, and

|H\cap aKa^{-1}|=2

. Find

|HaK|

. 2. Let

H,K

be finite subgroups of a group

G

. Suppose

|H|=9

|K|=4

, and

H\cap aKa^{-1}=\{e\}

. Find

|HaK|

. 3. Let

(G,\circ)

be a group and let

K

be a subgroup of

G

. Prove that

aKa^{-1}

and

K

have the same cardinality.

1. By the double coset cardinality formula,

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

2. Since

H\cap aKa^{-1}=\{e\}

, we get

\begin{equation*} |HaK|=|H||K|=9\cdot 4=36. \end{equation*}

3. Define

f:K\to aKa^{-1}

\begin{equation*} f(k)=a\circ k\circ a^{-1}. \end{equation*}

f(k_1)=f(k_2)

, then

\begin{align*} a\circ k_1\circ a^{-1}=a\circ k_2\circ a^{-1} &\implies k_1=k_2. \end{align*}

Thus

f

is one-to-one. If

x\in aKa^{-1}

, then

x=a\circ k\circ a^{-1}

for some

k\in K

, so

x=f(k)

. Thus

f

is onto. Therefore

K

and

aKa^{-1}

have the same cardinality.

This intersection counts exactly how many pairs in

H\times K

represent the same element of

HaK

It is a related counting formula. When

a=e

, it becomes

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

No. It holds when

H\cap aKa^{-1}

is the trivial subgroup.

BMLabs Mathematics

BMLabs Mathematics

Cardinality of Double Cosets

Cardinality of Double Cosets

OVERVIEW

CONJUGATE SUBGROUP

EXPLORE WHY THE DENOMINATOR APPEARS

PREIMAGE COUNTING EXPLORER

CONNECT THE COUNTS

CARDINALITY IN AN ADDITIVE ABELIAN GROUP

USING THE DOUBLE COSET FORMULA

SOLUTION

CALCULATE THE CARDINALITY FORMULA

DOUBLE COSET CARDINALITY CALCULATOR

A TRIVIAL INTERSECTION CASE

SOLUTION

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation