Equal and Disjoint Cosets

We now study the most important structural fact about cosets. Two cosets of the same subgroup cannot partly overlap. If they share even one element, then they are exactly the same coset. If they do not share any element, then they are disjoint. This is the reason cosets behave like separate blocks inside a group. Students often expect two translated copies of a subgroup to overlap partially, but group inverses prevent that from happening.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G-H

, then

\begin{equation*} aH\cap H=\varnothing. \end{equation*}

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G-H

, then

\begin{equation*} aH\cap H=\varnothing. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a\in G-H

. To prove that

aH\cap H=\varnothing

. If possible let

\begin{equation*} aH\cap H\neq\varnothing. \end{equation*}

Then there exists

x\in G

such that

\begin{equation*} x\in aH\cap H. \end{equation*}

Therefore

x\in aH

and

x\in H

. Since

x\in aH

, there exists

h\in H

such that

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

Then

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

Since

x\in H

and

h^{-1}\in H

, we get

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

. Therefore

a\in H

. This contradicts

a\in G-H

. Hence,

\begin{equation*} aH\cap H=\varnothing. \end{equation*}

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G-H

, then

\begin{equation*} Ha\cap H=\varnothing. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a\in G-H

. To prove that

Ha\cap H=\varnothing

. If possible let

\begin{equation*} Ha\cap H\neq\varnothing. \end{equation*}

Then there exists

x\in G

such that

\begin{equation*} x\in Ha\cap H. \end{equation*}

Therefore

x\in Ha

and

x\in H

. Since

x\in Ha

, there exists

h\in H

such that

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

Then

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

Since

h^{-1}\in H

and

x\in H

, we get

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

. Therefore

a\in H

. This contradicts

a\in G-H

. Hence,

\begin{equation*} Ha\cap H=\varnothing. \end{equation*}

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a,b\in G

, then either

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

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

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a,b\in G

. To prove that either

aH=bH

aH\cap bH=\varnothing

. Assume that

\begin{equation*} aH\cap bH\neq\varnothing. \end{equation*}

To prove that

aH=bH

. Let

x\in aH\cap bH

. Then there exist

h_1,h_2\in H

such that

\begin{equation*} x=a\circ h_1=b\circ h_2. \end{equation*}

Therefore

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

Let

y\in aH

. Then there exists

h\in H

such that

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

Using

a=b\circ (h_2\circ h_1^{-1})

, we get

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

Since

h_2,h_1^{-1},h\in H

, we get

h_2\circ h_1^{-1}\circ h\in H

. Therefore

y\in bH

. Therefore

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

Similarly, by interchanging

a

and

b

, we get

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

Therefore

aH=bH

. Hence, either

aH=bH

aH\cap bH=\varnothing

\square

We observe the equal-or-disjoint theorem by comparing two cosets in

\mathbb{Z}_n

. Choose two representatives and watch the intersection of the two cosets. If the intersection has even one element, the two cosets become exactly equal. If the cosets are not equal, their intersection is empty, so partial overlap never occurs.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a,b\in G

, then either

\begin{equation*} Ha=Hb \end{equation*}

\begin{equation*} Ha\cap Hb=\varnothing. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a,b\in G

. To prove that either

Ha=Hb

Ha\cap Hb=\varnothing

. Assume that

\begin{equation*} Ha\cap Hb\neq\varnothing. \end{equation*}

To prove that

Ha=Hb

. Let

x\in Ha\cap Hb

. Then there exist

h_1,h_2\in H

such that

\begin{equation*} x=h_1\circ a=h_2\circ b. \end{equation*}

Therefore

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

Let

y\in Ha

. Then there exists

h\in H

such that

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

Using

a=(h_1^{-1}\circ h_2)\circ b

, we get

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

Since

h,h_1^{-1},h_2\in H

, we get

h\circ h_1^{-1}\circ h_2\in H

. Therefore

y\in Hb

. Therefore

\begin{equation*} Ha\subseteq Hb. \end{equation*}

Similarly, by interchanging

a

and

b

, we get

\begin{equation*} Hb\subseteq Ha. \end{equation*}

Therefore

Ha=Hb

. Hence, either

Ha=Hb

Ha\cap Hb=\varnothing

\square

Let

G=\mathbb{Z}_{12}

under addition modulo

12

and let

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

The distinct cosets of

H

are

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

Any two of these cosets are disjoint. For example,

\begin{equation*} (1+H)\cap (2+H)=\varnothing. \end{equation*}

Also,

\begin{equation*} 4+H=H, \end{equation*}

because

4\in H

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Suppose

aH\cap bH\neq\varnothing

. Prove that

b\in aH

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Given that

aH\cap bH\neq\varnothing

. By the equal or disjoint theorem for left cosets, we get

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

Since

e\in H

, we get

\begin{align*} b &=b\circ e\\ &\in bH\\ &=aH. \end{align*}

Therefore

b\in aH

We compute two modular cosets and their intersection directly. This lets us test the equal-or-disjoint theorem without relying on a diagram. If the intersection is nonempty, the calculator reports equality; otherwise it reports disjointness. The result should never be a nonempty proper overlap.

1. Let

G=\mathbb{Z}_{9}

under addition modulo

9

and let

H=\{0,3,6\}

. List all distinct cosets of

H

. 2. Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

aH\cap H\neq\varnothing

, prove that

a\in H

. 3. Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

Ha\cap Hb\neq\varnothing

, prove that

Ha=Hb

1. The distinct cosets are

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

2. Since

aH\cap H\neq\varnothing

and

H=eH

, the equal or disjoint theorem gives

aH=H

. Hence,

a\in H

. 3. This is exactly the right coset equal or disjoint theorem. Since the intersection is nonempty, the disjoint alternative is impossible. Therefore

Ha=Hb

No. If two cosets of the same subgroup have one common element, then they are equal.

It shows that cosets divide the group into non-overlapping blocks.

Yes. The proof is similar, but multiplication must be placed on the correct side.

BMLabs Mathematics

BMLabs Mathematics

Equal and Disjoint Cosets

Equal and Disjoint Cosets

OVERVIEW

EXPLORE EQUAL OR DISJOINT COSETS

EQUAL OR DISJOINT COSETS EXPLORER

DISJOINT COSETS IN INTEGERS MODULO TWELVE

USING ONE COMMON ELEMENT

SOLUTION

TEST COSET INTERSECTIONS

COSET INTERSECTION TESTER

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation