Cardinality and Correspondence of Cosets

We now prove that all cosets of the same subgroup have the same number of elements. This fact is not only a counting statement; it is the mechanism behind Lagrange's theorem. Each coset is a translated copy of the subgroup, so it has the same size as the subgroup. We also prove that the collection of left cosets and the collection of right cosets have the same cardinality. This does not mean each left coset equals a right coset, but it does mean they can be paired perfectly.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then any two left cosets of

H

G

have the same cardinality.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then any two left cosets of

H

G

have the same cardinality.

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that any two left cosets of

H

G

have the same cardinality. Let

a,b\in G

. We prove that

aH

and

bH

have the same cardinality. Define

f:aH\to bH

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

for every

h\in H

. First, we prove that

f

is well-defined. Let

a\circ h_1=a\circ h_2

, where

h_1,h_2\in H

. Then

\begin{align*} a\circ h_1=a\circ h_2 &\implies a^{-1}\circ a\circ h_1=a^{-1}\circ a\circ h_2\\ &\implies h_1=h_2. \end{align*}

Therefore

b\circ h_1=b\circ h_2

, so

f

is well-defined. To prove that

f

is one-one, let

f(a\circ h_1)=f(a\circ h_2)

. Then

\begin{align*} f(a\circ h_1)=f(a\circ h_2) &\implies b\circ h_1=b\circ h_2\\ &\implies b^{-1}\circ b\circ h_1=b^{-1}\circ b\circ h_2\\ &\implies h_1=h_2\\ &\implies a\circ h_1=a\circ h_2. \end{align*}

Therefore

f

is one-one. To prove that

f

is onto, let

y\in bH

. Then there exists

h\in H

such that

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

Now

a\circ h\in aH

and

\begin{equation*} f(a\circ h)=b\circ h=y. \end{equation*}

Therefore

f

is onto. Hence,

aH

and

bH

have the same cardinality.

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G

, then

\begin{equation*} |aH|=|H|. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a\in G

. To prove that

|aH|=|H|

. Since

H=eH

, the previous theorem gives

\begin{equation*} |aH|=|eH|. \end{equation*}

Since

eH=H

, we get

\begin{equation*} |aH|=|H|. \end{equation*}

Hence,

|aH|=|H|

\square

We observe that translating a subgroup does not change the number of elements. Choose a subgroup of

\mathbb{Z}_n

and a representative, and the preview pairs each element of

H

with the corresponding element of

a+H

. The values may change, but the number of entries stays the same. This is the finite counting idea behind Lagrange's theorem.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then any two right cosets of

H

G

have the same cardinality.

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that any two right cosets of

H

G

have the same cardinality. Let

a,b\in G

. We prove that

Ha

and

Hb

have the same cardinality. Define

g:Ha\to Hb

\begin{equation*} g(h\circ a)=h\circ b \end{equation*}

for every

h\in H

. First, we prove that

g

is well-defined. Let

h_1\circ a=h_2\circ a

, where

h_1,h_2\in H

. Then

\begin{align*} h_1\circ a=h_2\circ a &\implies h_1\circ a\circ a^{-1}=h_2\circ a\circ a^{-1}\\ &\implies h_1=h_2. \end{align*}

Therefore

h_1\circ b=h_2\circ b

, so

g

is well-defined. To prove that

g

is one-one, let

g(h_1\circ a)=g(h_2\circ a)

. Then

\begin{align*} g(h_1\circ a)=g(h_2\circ a) &\implies h_1\circ b=h_2\circ b\\ &\implies h_1\circ b\circ b^{-1}=h_2\circ b\circ b^{-1}\\ &\implies h_1=h_2\\ &\implies h_1\circ a=h_2\circ a. \end{align*}

Therefore

g

is one-one. To prove that

g

is onto, let

y\in Hb

. Then there exists

h\in H

such that

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

Now

h\circ a\in Ha

and

\begin{equation*} g(h\circ a)=h\circ b=y. \end{equation*}

Therefore

g

is onto. Hence,

Ha

and

Hb

have the same cardinality.

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G

, then

\begin{equation*} |Ha|=|H|. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a\in G

. To prove that

|Ha|=|H|

. Since

H=He

, the previous theorem gives

\begin{equation*} |Ha|=|He|. \end{equation*}

Since

He=H

, we get

\begin{equation*} |Ha|=|H|. \end{equation*}

Hence,

|Ha|=|H|

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then there is a one-one correspondence from the set of all left cosets of

H

G

onto the set of all right cosets of

H

G

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that there is a one-one correspondence from the set of all left cosets of

H

onto the set of all right cosets of

H

. Let

\mathcal{L}

be the set of all left cosets of

H

and let

\mathcal{R}

be the set of all right cosets of

H

. Define

\Phi:\mathcal{L}\to \mathcal{R}

\begin{equation*} \Phi(aH)=Ha^{-1}. \end{equation*}

First, we prove that

\Phi

is well-defined. Let

aH=bH

. By the left coset equality criterion,

\begin{equation*} b^{-1}\circ a\in H. \end{equation*}

For right cosets, we have

\begin{align*} Ha^{-1}=Hb^{-1} &\iff b^{-1}\circ a\in H. \end{align*}

Therefore

Ha^{-1}=Hb^{-1}

. Hence,

\Phi

is well-defined. To prove that

\Phi

is one-one, let

\begin{equation*} \Phi(aH)=\Phi(bH). \end{equation*}

Then

\begin{equation*} Ha^{-1}=Hb^{-1}. \end{equation*}

By the right coset equality criterion,

\begin{equation*} b^{-1}\circ a\in H. \end{equation*}

By the left coset equality criterion,

\begin{equation*} aH=bH. \end{equation*}

Therefore

\Phi

is one-one. To prove that

\Phi

is onto, let

Ha\in \mathcal{R}

. Then

a^{-1}H\in \mathcal{L}

and

\begin{align*} \Phi(a^{-1}H) &=H(a^{-1})^{-1}\\ &=Ha. \end{align*}

Therefore

\Phi

is onto. Hence, there is a one-one correspondence from the set of all left cosets of

H

onto the set of all right cosets of

H

\square

Let

G=\mathbb{Z}_{12}

and let

H=\{0,4,8\}

. Each coset of

H

has

3

elements:

\begin{align*} 0+H&=\{0,4,8\},\\ 1+H&=\{1,5,9\},\\ 2+H&=\{2,6,10\},\\ 3+H&=\{3,7,11\}. \end{align*}

Therefore every coset has the same cardinality as

H

Let

(G,\circ)

be a group and let

H

be a finite subgroup of

G

with

|H|=6

. If

a\in G

, find

|aH|

and

|Ha|

Let

(G,\circ)

be a group and let

H

be a finite subgroup of

G

with

|H|=6

. Since every left coset has the same cardinality as

H

, we get

\begin{equation*} |aH|=|H|=6. \end{equation*}

Since every right coset has the same cardinality as

H

, we get

\begin{equation*} |Ha|=|H|=6. \end{equation*}

We compute the size of a modular subgroup and one of its cosets. The list of elements changes after translation, but the number of elements does not. This is because the map

h\mapsto a+h

pairs every subgroup element with exactly one coset element. The calculator verifies the equality

|a+H|=|H|

in concrete examples.

1. Let

H

be a finite subgroup of a group

G

and let

|H|=4

. Find the number of elements in

aH

for any

a\in G

. 2. Let

G=\mathbb{Z}_{18}

and let

H=\{0,6,12\}

. Find the number of elements in each coset of

H

. 3. Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Describe a one-one correspondence from left cosets of

H

to right cosets of

H

1. We have

\begin{equation*} |aH|=|H|=4. \end{equation*}

2. Since

|H|=3

, each coset of

H

has

3

elements. 3. The correspondence is

\begin{equation*} aH\mapsto Ha^{-1}. \end{equation*}

Yes. Both left and right cosets have the same cardinality as

H

No. A coset may have the same size as

H

without being equal to

H

Yes. The map

aH\mapsto Ha^{-1}

gives a one-one correspondence.

BMLabs Mathematics

BMLabs Mathematics

Cardinality and Correspondence of Cosets

Cardinality and Correspondence of Cosets

OVERVIEW

EXPLORE COSET CARDINALITY

COSET CARDINALITY EXPLORER

CARDINALITY OF COSETS IN INTEGERS MODULO TWELVE

COUNTING ELEMENTS IN A COSET

SOLUTION

CALCULATE COSET SIZE

COSET SIZE CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation