Cosets as Partitions

The equal-or-disjoint theorem has a powerful consequence: all left cosets of a subgroup form a partition of the group, and all right cosets also form a partition. A partition is a decomposition into nonempty, pairwise disjoint subsets whose union is the whole set. In this lecture, cosets move from being individual subsets to being a complete organizational system for the group. This viewpoint is essential for index and Lagrange's theorem, where we count the number of cosets and multiply by the size of one coset.

Let

X

be a nonempty set. A collection

\mathcal{P}

of nonempty subsets of

X

is called a

\textbf{partition}

X

if the following conditions hold: (i) every element of

X

belongs to at least one member of

\mathcal{P}

; (ii) any two distinct members of

\mathcal{P}

are disjoint.

Let

X

be a nonempty set. A collection

\mathcal{P}

of nonempty subsets of

X

is called a

\textbf{partition}

X

if the following conditions hold: (i) every element of

X

belongs to at least one member of

\mathcal{P}

; (ii) any two distinct members of

\mathcal{P}

are disjoint.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then the set of all left cosets of

H

G

forms a partition of

G

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that the set of all left cosets of

H

G

forms a partition of

G

. [1] To prove that every element of

G

belongs to at least one left coset of

H

. Let

a\in G

. Since

e\in H

, we get

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

Therefore every element of

G

belongs to at least one left coset of

H

. [2] To prove that any two distinct left cosets are disjoint. Let

aH

and

bH

be two left cosets of

H

G

. By the equal-or-disjoint theorem for left cosets, either

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

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

Therefore two distinct left cosets must be disjoint. Hence, the set of all left cosets of

H

G

forms a partition of

G

\square

We observe how the left cosets of a subgroup divide

\mathbb{Z}_n

into non-overlapping blocks. Choose a modulus and subgroup step, and the preview lists every distinct coset exactly once. Notice that every residue appears in one block and no residue appears in two different blocks. This is the partition idea that later supports index and Lagrange's theorem.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Then the set of all right cosets of

H

G

forms a partition of

G

Given that

(G,\circ)

is a group and

H

is a subgroup of

G

. To prove that the set of all right cosets of

H

G

forms a partition of

G

. [1] To prove that every element of

G

belongs to at least one right coset of

H

. Let

a\in G

. Since

e\in H

, we get

\begin{align*} a &=e\circ a\\ &\in Ha. \end{align*}

Therefore every element of

G

belongs to at least one right coset of

H

. [2] To prove that any two distinct right cosets are disjoint. Let

Ha

and

Hb

be two right cosets of

H

G

. By the equal-or-disjoint theorem for right cosets, either

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

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

Therefore two distinct right cosets must be disjoint. Hence, the set of all right cosets of

H

G

forms a partition of

G

\square

Let

G=\mathbb{Z}_8

under addition modulo

8

and let

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

The cosets of

H

are

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

These four subsets are nonempty, pairwise disjoint, and their union is

\mathbb{Z}_8

. Therefore they form a partition of

\mathbb{Z}_8

Let

G=\mathbb{Z}_8

and

H=\{0,4\}

. Then

\begin{align*} 1+H&=\{1,5\},\\ 5+H&=\{5,1\}. \end{align*}

Therefore

\begin{equation*} 1+H=5+H. \end{equation*}

Hence, the same coset may have more than one representative.

Let

G=\mathbb{Z}_{10}

under addition modulo

10

and let

H=\{0,5\}

. Find the partition of

G

formed by the cosets of

H

Let

G=\mathbb{Z}_{10}

and

H=\{0,5\}

. The cosets are

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

The remaining representatives repeat these cosets:

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

Therefore the required partition is

\begin{equation*} \{\{0,5\},\{1,6\},\{2,7\},\{3,8\},\{4,9\}\}. \end{equation*}

We compute all distinct cosets of a subgroup of

\mathbb{Z}_n

. The calculator checks repeated representatives and keeps only one copy of each coset. The number of listed cosets is the number of blocks in the partition. Since every block has the same size, this is the counting pattern that later becomes Lagrange's theorem.

1. Let

G=\mathbb{Z}_{12}

and let

H=\{0,6\}

. Find the partition of

G

formed by the cosets of

H

. 2. Let

G=\mathbb{Z}_{6}

and let

H=\{0,3\}

. Decide whether

\{0,3\}

and

\{1,4\}

can overlap. 3. Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Explain why every element of

G

belongs to some left coset of

H

1. The partition is

\begin{equation*} \{\{0,6\},\{1,7\},\{2,8\},\{3,9\},\{4,10\},\{5,11\}\}. \end{equation*}

2. They cannot overlap. They are distinct cosets of the same subgroup, so they are disjoint. 3. If

a\in G

, then

a=a\circ e

and

e\in H

. Therefore

a\in aH

The union is the whole group

G

It may appear in two listed cosets only when those cosets are actually equal.

It lets us count a group by counting cosets and the number of elements in each coset.

BMLabs Mathematics

BMLabs Mathematics

Cosets as Partitions

Cosets as Partitions

OVERVIEW

PARTITION

EXPLORE THE COSET PARTITION

COSET PARTITION EXPLORER

PARTITION OF INTEGERS MODULO EIGHT

REPRESENTATIVES ARE NOT UNIQUE

FINDING A COSET PARTITION

SOLUTION

CALCULATE A FULL COSET PARTITION

COSET PARTITION CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation