Cosets and Lagrange's Theorem

Beginning with the definition of cosets of a subgroup, we prove Lagrange's theorem --- the assertion that the order of a subgroup divides the order of the group. As applications, we derive Fermat's little theorem and Euler's theorem, and we exhibit the alternating group A_4 as a counterexample to the converse.

Folio Official

March 1, 2026

1 Introduction

If $H$ is a subgroup of a finite group $G$ , what can we say about the relationship between $∣ H ∣$ and $∣ G ∣$ ? Consider, for instance, $S_{3}$ , which has order $6$ . Its subgroups have orders $1, 2, 3,$ and $6$ , each of which divides $6$ . This is no coincidence.

Lagrange's theorem establishes this divisibility in full generality. The key idea is the notion of a coset.

2 Definition of Cosets

Definition 1 (Left and right cosets).

Let

G

be a group,

H \leq G

, and

a \in G

. The sets

a H H a = {ah ∣ h \in H} (the left coset of H containing a) = {ha ∣ h \in H} (the right coset of H containing a)

are called the left and right cosets of

H

G

determined by

a

Example 2.

Let

G = S_{3}

and

H = {e, (12)}

, a subgroup of order

2

. The left cosets of

H

are:

eH = (12) H (13) H = (123) H (23) H = (132) H = {e, (12)} = {(13), (123)} = {(23), (132)}

These three cosets partition

S_{3}

Example 3.

Let

G = (Z, +)

and

H = 3 Z

. The cosets (written additively as

a + H

) are:

0 + 3 Z 1 + 3 Z 2 + 3 Z = {\dots, - 6, - 3, 0, 3, 6, \dots} = {\dots, - 5, - 2, 1, 4, 7, \dots} = {\dots, - 4, - 1, 2, 5, 8, \dots}

This is simply the classification of integers by their remainder upon division by

3

— the origin of the term "residue class."

3 Basic Properties of Cosets

Theorem 4 (Cosets partition the group).

Let

H \leq G

. The left cosets of

H

G

satisfy the following:

Every element belongs to some coset: $a \in a H$ for all $a \in G$ .
$a H = b H$ if and only if $a^{- 1} b \in H$ .
Distinct cosets are disjoint: $a H \neq = b H$ implies $a H \cap b H = \emptyset$ .
All left cosets have the same cardinality: $∣ a H ∣ = ∣ H ∣$ .

Consequently, the left cosets of

H

form a partition of

G

Proof.

(1)

a = a e \in a H

(2) If

a H = b H

, then

b = b e \in b H = a H

, so

b = ah

for some

h \in H

. Hence

a^{- 1} b = h \in H

. Conversely, if

a^{- 1} b = h \in H

, then

b = ah

, and for any

b h^{'} \in b H

we have

b h^{'} = ah h^{'} \in a H

(since

h h^{'} \in H

), giving

b H \subseteq a H

. Similarly,

a = b h^{- 1}

implies

a H \subseteq b H

(3) Suppose

a H \cap b H \neq = \emptyset

, and pick

c \in a H \cap b H

. Then

c = a h_{1} = b h_{2}

for some

h_{1}, h_{2} \in H

, so

a^{- 1} b = h_{1} h_{2}^{- 1} \in H

. By part (2),

a H = b H

. Taking the contrapositive gives (3).

(4) The map

f : H \to a H

defined by

f (h) = ah

is a bijection (surjectivity is by definition; injectivity follows from the cancellation law). □

Definition 5 (Formulation as an equivalence relation).

For

H \leq G

, define a relation

\sim

G

a \sim b ⟺ a^{- 1} b \in H .

Then

\sim

is an equivalence relation, and the equivalence class of

a

a H

Proof.

Reflexivity:

a^{- 1} a = e \in H

. Symmetry:

a^{- 1} b \in H \Rightarrow (a^{- 1} b)^{- 1} = b^{- 1} a \in H

. Transitivity:

a^{- 1} b, b^{- 1} c \in H \Rightarrow a^{- 1} c = (a^{- 1} b) (b^{- 1} c) \in H

. □

4 Lagrange's Theorem

Definition 6 (Index).

For

H \leq G

, the number of left cosets of

H

G

is called the index of

H

G

and is denoted

[G : H]

Theorem 7 (Lagrange's theorem).

Let

G

be a finite group and

H \leq G

. Then

∣ G ∣ = [G : H] \cdot ∣ H ∣.

In particular,

∣ H ∣

divides

∣ G ∣

Proof.

Let the distinct left cosets of

H

a_{1} H, a_{2} H, \dots, a_{k} H

, where

k = [G : H]

. These cosets partition

G

G = a_{1} H ⊔ a_{2} H ⊔ \dots ⊔ a_{k} H (disjoint union).

Since each coset has cardinality

∣ H ∣

∣ G ∣ = k \cdot ∣ H ∣ = [G : H] \cdot ∣ H ∣.

□

Remark 8.

The essence of the proof is that

G

can be sliced into copies of

H

. Each slice has size

∣ H ∣

, and the number of slices is

[G : H]

, so the total size is their product.

5 Corollaries of Lagrange's Theorem

Lagrange's theorem has numerous important consequences.

Theorem 9 (Order of an element divides the order of the group).

For an element

a

of a finite group

G

ord (a)

divides

∣ G ∣

. In particular,

a^{∣ G ∣} = e

Proof.

We have

ord (a) = ∣ ⟨ a ⟩ ∣

and

⟨ a ⟩ \leq G

, so Lagrange's theorem gives

ord (a) ∣ ∣ G ∣

. Writing

∣ G ∣ = ord (a) \cdot m

, we find

a^{∣ G ∣} = (a^{ord (a)})^{m} = e^{m} = e

. □

Theorem 10 (Groups of prime order are cyclic).

∣ G ∣ = p

where

p

is prime, then

G

is cyclic and

G ≅ Z / p Z

Proof.

Pick any

a \in G

with

a \neq = e

. Since

ord (a)

divides

∣ G ∣ = p

and

ord (a) \neq = 1

(because

a \neq = e

), we must have

ord (a) = p

. Therefore

⟨ a ⟩ = G

. □

Remark 11.

This is a powerful result: the structure of a group of prime order is completely determined. Groups of order

2, 3, 5, 7, 11, \dots

are all cyclic, and in each case there is essentially only one group (up to isomorphism).

Theorem 12 (Fermat's little theorem).

Let

p

be a prime and let

a

be an integer with

g cd (a, p) = 1

. Then

a^{p - 1} \equiv 1 (mod p)

Proof.

The set

(Z / p Z)^{\times} = {\overset{ˉ}{1}, \overset{ˉ}{2}, \dots, \overline{p - 1}}

is a group of order

p - 1

under multiplication (since

p

is prime, every nonzero element has a multiplicative inverse). Since

g cd (a, p) = 1

, we have

\overset{a}{ˉ} \in (Z / p Z)^{\times}

. By the corollary to Lagrange's theorem,

\overset{a}{ˉ}^{p - 1} = \overset{ˉ}{1}

, i.e.

a^{p - 1} \equiv 1 (mod p)

. □

Theorem 13 (Euler's theorem).

Let

n

be a positive integer and let

a

be an integer with

g cd (a, n) = 1

. Then

a^{φ (n)} \equiv 1 (mod n)

, where

φ (n)

denotes Euler's totient function (the number of integers between

1

and

n

coprime to

n

Proof.

The set

(Z / n Z)^{\times} = {\overset{a}{ˉ} ∣ g cd (a, n) = 1}

is a group of order

φ (n)

under multiplication. By the corollary to Lagrange's theorem,

\overset{a}{ˉ}^{φ (n)} = \overset{ˉ}{1}

. □

Remark 14.

Fermat's little theorem is the special case

n = p

(a prime) of Euler's theorem, since

φ (p) = p - 1

. The fact that these classical results from number theory both follow from the single group-theoretic statement of Lagrange's theorem is a striking demonstration of the power of abstract algebra.

6 The Converse of Lagrange's Theorem Fails

Theorem 15 (A counterexample to the converse).

d ∣ ∣ G ∣

, the group

G

need not possess a subgroup of order

d

Example 16.

Consider the alternating group

A_{4}

of order

12

. We have

6 ∣ 12

, yet

A_{4}

has no subgroup of order

6

To see this, we enumerate the elements of

A_{4}

The identity $e$ ( $1$ element).
The $3$ -cycles $(ab c)$ ( $8$ elements): $(123), (132), (124), (142), (134), (143), (234), (243)$ .
Products of two disjoint transpositions $(ab) (c d)$ ( $3$ elements): $(12) (34), (13) (24), (14) (23)$ .

The total is

12 = 4! /2

Suppose a subgroup

H

of order

6

exists. Since

[A_{4} : H] = 2

, the subgroup

H

is normal in

A_{4}

(a subgroup of index

2

is always normal, as we shall prove in the next chapter).

Since

H

is normal, it must be a union of conjugacy classes of

A_{4}

. The conjugacy classes of

A_{4}

are:

{e} (1), {(12) (34), (13) (24), (14) (23)} (3), {(123), (124), (134), (234)} (4), {(132), (142), (143), (243)} (4) .

The sizes are

1, 3, 4, 4

. To get

∣ H ∣ = 6

, we would need a sub-sum of

1 + 3 + 4 + 4

equalling

6

. But

1 + 3 = 4

1 + 4 = 5

1 + 3 + 4 = 8

— none give

6

. This is a contradiction.

Remark 17.

For finite abelian groups, the converse does hold: if

d ∣ ∣ G ∣

, then

G

has a subgroup of order

d

(a consequence of the structure theorem for finite abelian groups). Moreover, the Sylow theorems guarantee the existence of subgroups of order

p^{k}

whenever

p^{k} ∣ ∣ G ∣

, even for non-abelian groups.

7 Theorems on the Index

Theorem 18 (Multiplicativity of the index).

K \leq H \leq G

and

[G : K]

is finite, then

[G : K] = [G : H] \cdot [H : K] .

Proof.

Let

a_{1} H, \dots, a_{m} H

(

m = [G : H]

) be the left cosets of

H

G

, and let

b_{1} K, \dots, b_{n} K

(

n = [H : K]

) be the left cosets of

K

H

. Then

G = i ⨆ a_{i} H = i ⨆ a_{i} j ⨆ b_{j} K = i, j ⨆ a_{i} b_{j} K,

{a_{i} b_{j} K}

is the complete set of left cosets of

K

G

, giving

[G : K] = mn

To confirm that the cosets

a_{i} b_{j} K

are all distinct: if

a_{i} b_{j} K = a_{i^{'}} b_{j^{'}} K

, then

a_{i} b_{j}

and

a_{i^{'}} b_{j^{'}}

lie in the same

K

-coset. Thus

a_{i} b_{j} b_{j^{'}}^{- 1} a_{i^{'}}^{- 1} \in K \subseteq H

, which gives

a_{i} H = a_{i^{'}} H

, so

i = i^{'}

. It then follows that

b_{j} K = b_{j^{'}} K

, so

j = j^{'}

. □

Theorem 19 (Index of an intersection).

Let

H, K \leq G

with

[G : H]

and

[G : K]

both finite. Then

[G : H \cap K] \leq [G : H] \cdot [G : K] .

Equality holds when

[G : H]

and

[G : K]

are coprime.

Proof.

The map

ψ : G / (H \cap K) \to G / H \times G / K

defined by

a (H \cap K) \mapsto (a H, a K)

is well-defined and injective (if

a H = b H

and

a K = b K

, then

a^{- 1} b \in H \cap K

). So

[G : H \cap K] \leq [G : H] [G : K]

When

g cd ([G : H], [G : K]) = 1

: from

H \cap K \leq H

and

H \cap K \leq K

, multiplicativity of the index gives

[G : H] ∣ [G : H \cap K]

and

[G : K] ∣ [G : H \cap K]

. Since these indices are coprime,

[G : H] [G : K] ∣ [G : H \cap K]

. Combined with the reverse inequality, equality follows. □

8 Right Cosets and Double Cosets

Remark 20 (Left versus right).

In general,

a H \neq = H a

: left and right cosets may differ. However, the number of left cosets always equals the number of right cosets.

Theorem 21.

[G : H]_{L} = [G : H]_{R}

(the number of left cosets equals the number of right cosets).

Proof.

The map

a H \mapsto H a^{- 1}

is a bijection from the set of left cosets to the set of right cosets. Indeed,

a H = b H ⟺ a^{- 1} b \in H ⟺ b^{- 1} a \in H ⟺ H a^{- 1} = H b^{- 1}

, so the map is well-defined and injective. Surjectivity is clear. □

Definition 22 (Normal subgroups (preview)).

A subgroup

H \leq G

is called a normal subgroup, written

H ⊴ G

, if

a H = H a

for all

a \in G

. When

H

is normal, left cosets and right cosets coincide, and the set of cosets can be endowed with a natural group structure (the quotient group).

9 Worked Examples

Example 23 (

D_{4}

and

⟨ r^{2} ⟩

Consider the dihedral group

D_{4} = {e, r, r^{2}, r^{3}, s, rs, r^{2} s, r^{3} s}

of order

8

, and its subgroup

H = ⟨ r^{2} ⟩ = {e, r^{2}}

of order

2

. The left cosets are:

eH sH = {e, r^{2}}, = {s, r^{2} s}, rH rsH = {r, r^{3}}, = {rs, r^{3} s} .

There are

4

cosets, so

[D_{4} : H] = 4

. Check:

∣ D_{4} ∣ = 8 = 4 \cdot 2 = [D_{4} : H] \cdot ∣ H ∣

Example 24 (

S_{4}

and

A_{4}

Since

A_{4} \leq S_{4}

with

∣ A_{4} ∣ = 12

and

∣ S_{4} ∣ = 24

, we have

[S_{4} : A_{4}] = 2

. The two cosets are

A_{4}

(the even permutations) and

(12) A_{4}

(the odd permutations).

Example 25 (

Z

and

n Z

[Z : n Z] = n

. The cosets are

0 + n Z, 1 + n Z, \dots, (n - 1) + n Z

, and they form the additive group

Z / n Z

10 Summary and Next Steps

In this chapter we have covered:

The definition of left and right cosets, and the fact that they partition the group.
Lagrange's theorem: $∣ G ∣ = [G : H] \cdot ∣ H ∣$ .
Corollaries: the order of an element divides the order of the group; groups of prime order are cyclic.
Applications: Fermat's little theorem and Euler's theorem.
The converse of Lagrange's theorem fails in general (the $A_{4}$ counterexample).
Multiplicativity of the index and the index of an intersection.

Lagrange's theorem tells us that the order of a subgroup is constrained, but it does not address whether a subgroup of every allowable order actually exists. Answering this converse question (at least partially) is the goal of the Sylow theorems, which require the theory of normal subgroups and quotient groups as a prerequisite.

Group Theory Algebra Textbook Cosets Lagrange's Theorem

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Cosets and Lagrange's Theorem

Folio Official

March 1, 2026

1 Introduction

Lagrange's theorem establishes this divisibility in full generality. The key idea is the notion of a coset.

2 Definition of Cosets

Definition 1 (Left and right cosets).

Let

G

be a group,

H \leq G

, and

a \in G

. The sets

a H H a = {ah ∣ h \in H} (the left coset of H containing a) = {ha ∣ h \in H} (the right coset of H containing a)

are called the left and right cosets of

H

G

determined by

a

Example 2.

Let

G = S_{3}

and

H = {e, (12)}

, a subgroup of order

2

. The left cosets of

H

are:

eH = (12) H (13) H = (123) H (23) H = (132) H = {e, (12)} = {(13), (123)} = {(23), (132)}

These three cosets partition

S_{3}

Example 3.

Let

G = (Z, +)

and

H = 3 Z

. The cosets (written additively as

a + H

) are:

0 + 3 Z 1 + 3 Z 2 + 3 Z = {\dots, - 6, - 3, 0, 3, 6, \dots} = {\dots, - 5, - 2, 1, 4, 7, \dots} = {\dots, - 4, - 1, 2, 5, 8, \dots}

This is simply the classification of integers by their remainder upon division by

3

— the origin of the term "residue class."

3 Basic Properties of Cosets

Theorem 4 (Cosets partition the group).

Let

H \leq G

. The left cosets of

H

G

satisfy the following:

Every element belongs to some coset: $a \in a H$ for all $a \in G$ .
$a H = b H$ if and only if $a^{- 1} b \in H$ .
Distinct cosets are disjoint: $a H \neq = b H$ implies $a H \cap b H = \emptyset$ .
All left cosets have the same cardinality: $∣ a H ∣ = ∣ H ∣$ .

Consequently, the left cosets of

H

form a partition of

G

Proof.

(1)

a = a e \in a H

(2) If

a H = b H

, then

b = b e \in b H = a H

, so

b = ah

for some

h \in H

. Hence

a^{- 1} b = h \in H

. Conversely, if

a^{- 1} b = h \in H

, then

b = ah

, and for any

b h^{'} \in b H

we have

b h^{'} = ah h^{'} \in a H

(since

h h^{'} \in H

), giving

b H \subseteq a H

. Similarly,

a = b h^{- 1}

implies

a H \subseteq b H

(3) Suppose

a H \cap b H \neq = \emptyset

, and pick

c \in a H \cap b H

. Then

c = a h_{1} = b h_{2}

for some

h_{1}, h_{2} \in H

, so

a^{- 1} b = h_{1} h_{2}^{- 1} \in H

. By part (2),

a H = b H

. Taking the contrapositive gives (3).

(4) The map

f : H \to a H

defined by

f (h) = ah

is a bijection (surjectivity is by definition; injectivity follows from the cancellation law). □

Definition 5 (Formulation as an equivalence relation).

For

H \leq G

, define a relation

\sim

G

a \sim b ⟺ a^{- 1} b \in H .

Then

\sim

is an equivalence relation, and the equivalence class of

a

a H

Proof.

Reflexivity:

a^{- 1} a = e \in H

. Symmetry:

a^{- 1} b \in H \Rightarrow (a^{- 1} b)^{- 1} = b^{- 1} a \in H

. Transitivity:

a^{- 1} b, b^{- 1} c \in H \Rightarrow a^{- 1} c = (a^{- 1} b) (b^{- 1} c) \in H

. □

4 Lagrange's Theorem

Definition 6 (Index).

For

H \leq G

, the number of left cosets of

H

G

is called the index of

H

G

and is denoted

[G : H]

Theorem 7 (Lagrange's theorem).

Let

G

be a finite group and

H \leq G

. Then

∣ G ∣ = [G : H] \cdot ∣ H ∣.

In particular,

∣ H ∣

divides

∣ G ∣

Proof.

Let the distinct left cosets of

H

a_{1} H, a_{2} H, \dots, a_{k} H

, where

k = [G : H]

. These cosets partition

G

G = a_{1} H ⊔ a_{2} H ⊔ \dots ⊔ a_{k} H (disjoint union).

Since each coset has cardinality

∣ H ∣

∣ G ∣ = k \cdot ∣ H ∣ = [G : H] \cdot ∣ H ∣.

□

Remark 8.

The essence of the proof is that

G

can be sliced into copies of

H

. Each slice has size

∣ H ∣

, and the number of slices is

[G : H]

, so the total size is their product.

5 Corollaries of Lagrange's Theorem

Lagrange's theorem has numerous important consequences.

Theorem 9 (Order of an element divides the order of the group).

For an element

a

of a finite group

G

ord (a)

divides

∣ G ∣

. In particular,

a^{∣ G ∣} = e

Proof.

We have

ord (a) = ∣ ⟨ a ⟩ ∣

and

⟨ a ⟩ \leq G

, so Lagrange's theorem gives

ord (a) ∣ ∣ G ∣

. Writing

∣ G ∣ = ord (a) \cdot m

, we find

a^{∣ G ∣} = (a^{ord (a)})^{m} = e^{m} = e

. □

Theorem 10 (Groups of prime order are cyclic).

∣ G ∣ = p

where

p

is prime, then

G

is cyclic and

G ≅ Z / p Z

Proof.

Pick any

a \in G

with

a \neq = e

. Since

ord (a)

divides

∣ G ∣ = p

and

ord (a) \neq = 1

(because

a \neq = e

), we must have

ord (a) = p

. Therefore

⟨ a ⟩ = G

. □

Remark 11.

This is a powerful result: the structure of a group of prime order is completely determined. Groups of order

2, 3, 5, 7, 11, \dots

are all cyclic, and in each case there is essentially only one group (up to isomorphism).

Theorem 12 (Fermat's little theorem).

Let

p

be a prime and let

a

be an integer with

g cd (a, p) = 1

. Then

a^{p - 1} \equiv 1 (mod p)

Proof.

The set

(Z / p Z)^{\times} = {\overset{ˉ}{1}, \overset{ˉ}{2}, \dots, \overline{p - 1}}

is a group of order

p - 1

under multiplication (since

p

is prime, every nonzero element has a multiplicative inverse). Since

g cd (a, p) = 1

, we have

\overset{a}{ˉ} \in (Z / p Z)^{\times}

. By the corollary to Lagrange's theorem,

\overset{a}{ˉ}^{p - 1} = \overset{ˉ}{1}

, i.e.

a^{p - 1} \equiv 1 (mod p)

. □

Theorem 13 (Euler's theorem).

Let

n

be a positive integer and let

a

be an integer with

g cd (a, n) = 1

. Then

a^{φ (n)} \equiv 1 (mod n)

, where

φ (n)

denotes Euler's totient function (the number of integers between

1

and

n

coprime to

n

Proof.

The set

(Z / n Z)^{\times} = {\overset{a}{ˉ} ∣ g cd (a, n) = 1}

is a group of order

φ (n)

under multiplication. By the corollary to Lagrange's theorem,

\overset{a}{ˉ}^{φ (n)} = \overset{ˉ}{1}

. □

Remark 14.

Fermat's little theorem is the special case

n = p

(a prime) of Euler's theorem, since

φ (p) = p - 1

. The fact that these classical results from number theory both follow from the single group-theoretic statement of Lagrange's theorem is a striking demonstration of the power of abstract algebra.

6 The Converse of Lagrange's Theorem Fails

Theorem 15 (A counterexample to the converse).

d ∣ ∣ G ∣

, the group

G

need not possess a subgroup of order

d

Example 16.

Consider the alternating group

A_{4}

of order

12

. We have

6 ∣ 12

, yet

A_{4}

has no subgroup of order

6

To see this, we enumerate the elements of

A_{4}

The identity $e$ ( $1$ element).
The $3$ -cycles $(ab c)$ ( $8$ elements): $(123), (132), (124), (142), (134), (143), (234), (243)$ .
Products of two disjoint transpositions $(ab) (c d)$ ( $3$ elements): $(12) (34), (13) (24), (14) (23)$ .

The total is

12 = 4! /2

Suppose a subgroup

H

of order

6

exists. Since

[A_{4} : H] = 2

, the subgroup

H

is normal in

A_{4}

(a subgroup of index

2

is always normal, as we shall prove in the next chapter).

Since

H

is normal, it must be a union of conjugacy classes of

A_{4}

. The conjugacy classes of

A_{4}

are:

{e} (1), {(12) (34), (13) (24), (14) (23)} (3), {(123), (124), (134), (234)} (4), {(132), (142), (143), (243)} (4) .

The sizes are

1, 3, 4, 4

. To get

∣ H ∣ = 6

, we would need a sub-sum of

1 + 3 + 4 + 4

equalling

6

. But

1 + 3 = 4

1 + 4 = 5

1 + 3 + 4 = 8

— none give

6

. This is a contradiction.

Remark 17.

For finite abelian groups, the converse does hold: if

d ∣ ∣ G ∣

, then

G

has a subgroup of order

d

(a consequence of the structure theorem for finite abelian groups). Moreover, the Sylow theorems guarantee the existence of subgroups of order

p^{k}

whenever

p^{k} ∣ ∣ G ∣

, even for non-abelian groups.

7 Theorems on the Index

Theorem 18 (Multiplicativity of the index).

K \leq H \leq G

and

[G : K]

is finite, then

[G : K] = [G : H] \cdot [H : K] .

Proof.

Let

a_{1} H, \dots, a_{m} H

(

m = [G : H]

) be the left cosets of

H

G

, and let

b_{1} K, \dots, b_{n} K

(

n = [H : K]

) be the left cosets of

K

H

. Then

G = i ⨆ a_{i} H = i ⨆ a_{i} j ⨆ b_{j} K = i, j ⨆ a_{i} b_{j} K,

{a_{i} b_{j} K}

is the complete set of left cosets of

K

G

, giving

[G : K] = mn

To confirm that the cosets

a_{i} b_{j} K

are all distinct: if

a_{i} b_{j} K = a_{i^{'}} b_{j^{'}} K

, then

a_{i} b_{j}

and

a_{i^{'}} b_{j^{'}}

lie in the same

K

-coset. Thus

a_{i} b_{j} b_{j^{'}}^{- 1} a_{i^{'}}^{- 1} \in K \subseteq H

, which gives

a_{i} H = a_{i^{'}} H

, so

i = i^{'}

. It then follows that

b_{j} K = b_{j^{'}} K

, so

j = j^{'}

. □

Theorem 19 (Index of an intersection).

Let

H, K \leq G

with

[G : H]

and

[G : K]

both finite. Then

[G : H \cap K] \leq [G : H] \cdot [G : K] .

Equality holds when

[G : H]

and

[G : K]

are coprime.

Proof.

The map

ψ : G / (H \cap K) \to G / H \times G / K

defined by

a (H \cap K) \mapsto (a H, a K)

is well-defined and injective (if

a H = b H

and

a K = b K

, then

a^{- 1} b \in H \cap K

). So

[G : H \cap K] \leq [G : H] [G : K]

When

g cd ([G : H], [G : K]) = 1

: from

H \cap K \leq H

and

H \cap K \leq K

, multiplicativity of the index gives

[G : H] ∣ [G : H \cap K]

and

[G : K] ∣ [G : H \cap K]

. Since these indices are coprime,

[G : H] [G : K] ∣ [G : H \cap K]

. Combined with the reverse inequality, equality follows. □

8 Right Cosets and Double Cosets

Remark 20 (Left versus right).

In general,

a H \neq = H a

: left and right cosets may differ. However, the number of left cosets always equals the number of right cosets.

Theorem 21.

[G : H]_{L} = [G : H]_{R}

(the number of left cosets equals the number of right cosets).

Proof.

The map

a H \mapsto H a^{- 1}

is a bijection from the set of left cosets to the set of right cosets. Indeed,

a H = b H ⟺ a^{- 1} b \in H ⟺ b^{- 1} a \in H ⟺ H a^{- 1} = H b^{- 1}

, so the map is well-defined and injective. Surjectivity is clear. □

Definition 22 (Normal subgroups (preview)).

A subgroup

H \leq G

is called a normal subgroup, written

H ⊴ G

, if

a H = H a

for all

a \in G

. When

H

is normal, left cosets and right cosets coincide, and the set of cosets can be endowed with a natural group structure (the quotient group).

9 Worked Examples

Example 23 (

D_{4}

and

⟨ r^{2} ⟩

Consider the dihedral group

D_{4} = {e, r, r^{2}, r^{3}, s, rs, r^{2} s, r^{3} s}

of order

8

, and its subgroup

H = ⟨ r^{2} ⟩ = {e, r^{2}}

of order

2

. The left cosets are:

eH sH = {e, r^{2}}, = {s, r^{2} s}, rH rsH = {r, r^{3}}, = {rs, r^{3} s} .

There are

4

cosets, so

[D_{4} : H] = 4

. Check:

∣ D_{4} ∣ = 8 = 4 \cdot 2 = [D_{4} : H] \cdot ∣ H ∣

Example 24 (

S_{4}

and

A_{4}

Since

A_{4} \leq S_{4}

with

∣ A_{4} ∣ = 12

and

∣ S_{4} ∣ = 24

, we have

[S_{4} : A_{4}] = 2

. The two cosets are

A_{4}

(the even permutations) and

(12) A_{4}

(the odd permutations).

Example 25 (

Z

and

n Z

[Z : n Z] = n

. The cosets are

0 + n Z, 1 + n Z, \dots, (n - 1) + n Z

, and they form the additive group

Z / n Z

10 Summary and Next Steps

In this chapter we have covered:

The definition of left and right cosets, and the fact that they partition the group.
Lagrange's theorem: $∣ G ∣ = [G : H] \cdot ∣ H ∣$ .
Corollaries: the order of an element divides the order of the group; groups of prime order are cyclic.
Applications: Fermat's little theorem and Euler's theorem.
The converse of Lagrange's theorem fails in general (the $A_{4}$ counterexample).
Multiplicativity of the index and the index of an intersection.

Group Theory Algebra Textbook Cosets Lagrange's Theorem

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Cosets and Lagrange's Theorem

1 Introduction

2 Definition of Cosets

3 Basic Properties of Cosets

4 Lagrange's Theorem

5 Corollaries of Lagrange's Theorem

6 The Converse of Lagrange's Theorem Fails

7 Theorems on the Index

8 Right Cosets and Double Cosets

9 Worked Examples

10 Summary and Next Steps

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Cosets and Lagrange's Theorem

1 Introduction

2 Definition of Cosets

3 Basic Properties of Cosets

4 Lagrange's Theorem

5 Corollaries of Lagrange's Theorem

6 The Converse of Lagrange's Theorem Fails

7 Theorems on the Index

8 Right Cosets and Double Cosets

9 Worked Examples

10 Summary and Next Steps

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