Cosets Determined by Elements of a Subgroup

The previous lecture introduced left and right cosets. We now study the first important special case: what happens when the representative itself belongs to the subgroup. This case is foundational because it tells us exactly when a coset is not new at all. If the representative lies inside

H

, then multiplying

H

by that representative simply reproduces

H

. Conversely, if a coset determined by

h

equals

H

, then

h

must lie in

H

. This result is often the first place where students see that cosets behave like shifted copies of a subgroup.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

h\in G

, then

\begin{equation*} h\in H \iff hH=H. \end{equation*}

H

, then multiplying

H

by that representative simply reproduces

H

. Conversely, if a coset determined by

h

equals

H

, then

h

must lie in

H

. This result is often the first place where students see that cosets behave like shifted copies of a subgroup.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

h\in G

, then

\begin{equation*} h\in H \iff hH=H. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

h\in G

. To prove that

h\in H \iff hH=H

. [1] Let

h\in H

. To prove that

hH=H

. Let

x\in hH

. Then there exists

u\in H

such that

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

Since

h,u\in H

and

H

is closed under

\circ

, we get

x\in H

. Therefore

\begin{equation*} hH\subseteq H. \end{equation*}

Let

y\in H

. Since

h\in H

and

H

is a subgroup of

G

, we get

h^{-1}\in H

. Therefore

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

. Now

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

Therefore

y\in hH

. Therefore

\begin{equation*} H\subseteq hH. \end{equation*}

Therefore

\begin{equation*} hH=H. \end{equation*}

[2] Let

hH=H

. To prove that

h\in H

. Since

e\in H

, we get

\begin{align*} h &=h\circ e\\ &\in hH. \end{align*}

Since

hH=H

, we get

h\in H

. Hence,

\begin{equation*} h\in H \iff hH=H. \end{equation*}

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

h\in G

, then

\begin{equation*} h\in H \iff Hh=H. \end{equation*}

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

h\in G

. To prove that

h\in H \iff Hh=H

. [1] Let

h\in H

. To prove that

Hh=H

. Let

x\in Hh

. Then there exists

u\in H

such that

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

Since

u,h\in H

and

H

is closed under

\circ

, we get

x\in H

. Therefore

\begin{equation*} Hh\subseteq H. \end{equation*}

Let

y\in H

. Since

h\in H

and

H

is a subgroup of

G

, we get

h^{-1}\in H

. Therefore

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

. Now

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

Therefore

y\in Hh

. Therefore

\begin{equation*} H\subseteq Hh. \end{equation*}

Therefore

\begin{equation*} Hh=H. \end{equation*}

[2] Let

Hh=H

. To prove that

h\in H

. Since

e\in H

, we get

\begin{align*} h &=e\circ h\\ &\in Hh. \end{align*}

Since

Hh=H

, we get

h\in H

. Hence,

\begin{equation*} h\in H \iff Hh=H. \end{equation*}

\square

We observe the special case where the representative already belongs to the subgroup. Choose a subgroup of

\mathbb{Z}_n

and a representative, then compare

a+H

with

H

. When

a

lies in

H

, the translation only rearranges the same elements, so the coset is not new. When

a

does not lie in

H

, the coset is a shifted copy that differs from

H

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

a\in G-H

, then

\begin{equation*} aH\neq H \end{equation*}

and

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

Given that

(G,\circ)

is a group,

H

is a subgroup of

G

, and

a\in G-H

. To prove that

aH\neq H

and

Ha\neq H

. Since

a\in G-H

, we get

a\notin H

. If possible, let

aH=H

. By the previous theorem on left cosets,

\begin{equation*} aH=H \implies a\in H. \end{equation*}

A contradiction. Therefore

aH\neq H

. If possible, let

Ha=H

. By the previous theorem on right cosets,

\begin{equation*} Ha=H \implies a\in H. \end{equation*}

A contradiction. Hence,

aH\neq H

and

Ha\neq H

\square

Let

G=\mathbb{Z}_{12}

under addition modulo

12

and let

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

Since

3\in H

, the coset determined by

3

\begin{align*} 3+H &=\{3+0,\;3+3,\;3+6,\;3+9\}\\ &=\{3,\;6,\;9,\;12\}\\ &=\{0,3,6,9\}\\ &=H. \end{align*}

This example illustrates the theorem in additive notation.

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Suppose

aH=H

. Prove that

a^{-1}\in H

Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. Given that

aH=H

. Since

aH=H

, by the left coset criterion we get

\begin{equation*} a\in H. \end{equation*}

Since

H

is a subgroup of

G

, it is closed under taking inverses. Therefore

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

We test the theorem numerically in

\mathbb{Z}_n

. Enter a subgroup step and a representative, and the calculator checks both statements: whether the representative belongs to the subgroup and whether the coset equals the subgroup. Notice that the two answers always agree in this additive setting. This is the finite modular version of

h\in H

if and only if

hH=H

1. Let

G=\mathbb{Z}_{15}

under addition modulo

15

and let

H=\{0,5,10\}

. Find

10+H

. 2. Let

(G,\circ)

be a group and let

H

be a subgroup of

G

. If

Ha=H

, prove that

a^{-1}\in H

. 3. Let

H

be a subgroup of

G

. Prove that if

a\notin H

, then

aH

cannot be equal to

H

1. Since

10\in H

, we get

\begin{equation*} 10+H=H=\{0,5,10\}. \end{equation*}

2. Since

Ha=H

, the right coset criterion gives

a\in H

. Since

H

is a subgroup,

a^{-1}\in H

. 3. If

aH=H

, then the left coset criterion gives

a\in H

, which contradicts

a\notin H

. Therefore

aH\neq H

Multiplication by an element already inside

H

only rearranges the elements of

H

Yes. The theorem proves both directions:

hH=H

happens exactly when

h\in H

Yes. We also have

h\in H \iff Hh=H

BMLabs Mathematics

BMLabs Mathematics

Cosets Determined by Elements of a Subgroup

Cosets Determined by Elements of a Subgroup

OVERVIEW

EXPLORE WHEN A COSET IS THE SUBGROUP

COSET EQUALS SUBGROUP EXPLORER

COSETS DETERMINED BY ELEMENTS OF A SUBGROUP

TESTING WHETHER A COSET IS THE SUBGROUP

SOLUTION

TEST COSET SUBGROUP EQUALITY

COSET SUBGROUP EQUALITY TESTER

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation