SECTION 10 Cosets and the Theorem of Lagrange

Theorem

Let H be a subgroup of G. Let the relation L be defined on G by

a Lb if and only if a-1b H.

Let the relation R be defined on G by

a R b if and only if ab-1 H.

Then Land R both equivalence relations on G

Proof: exercise 26.

The equivalence relation L defines a partition of G, and the cell containing a is {ah | h H}, which we denote by aH. Similarly for the equivalence relation R.

Cosets

Definition

Let H be a subgroup of a group G. The subset aH = {ah | h H} of G is the left coset of H containing a, while the subset Ha = {ha | h H} is the right coset of H containing a.

Example

Example: Exhibit the left cosets and the right cosets of the subgroup 3Z of Z.

Solution: Our notation is additive, so the left coset of 3Z containing m is m+3Z.

• Let m=0, then 3Z = {…, -9, -6, -3, 0, 3, 6, 9, …} is one left coset.• Then select 1 in Z but not in 3Z, and find the left coset containing 1. We

have 1+3Z= {…, -8, -5, -2, 1, 4, 7, 10, …}.• Observe 3Z and 1+3Z do not yet exhaust Z, then select 2 in Z but not in

3Z or 1+3Z, and find the left coset containing 2. We have 2+3Z= {…, -7, -4, -1, 2, 5, 8, 11, …}

• Now 3Z, 1+3Z, and 2+3Z do exhaust Z, so these are all left cosest of 3Z.

Since Z is abelian, the let coset m+3Z and the right coset 3Z+m are the same.

Example

Fact:For a subgroup H of an abelian group G, the partition of G into left

cosets of H and the partition into the right cosets are the same. In this case, we do not have to specify left or right cosets.

Example: The group Z6 abelian. Find the partition of Z6 into cosets of the subgroup H={0, 3}.

Solution: • One coset is {0, 3} itself.• Select 1 in Z6 but not in {0, 3}. The coset containing 1 is 1+{0, 3} =

{1, 4}.• Similarly, select 2 in Z6 but not in {0, 3} or {1, 4}, the coset containing

2 is 2+{0, 3} = {2, 5}.• Since {0, 3}, {1, 4}, and {2, 5} exhaust all of Z6, these are all cosets.

Example

Let H be the subgroup <1>={0, 1} of S3. Find the partitions of S3 into left cosets of H, and the partition into right cosets of H. (See the table for the symmetric group S3 on the next slide.)

Solution: For the partition into left cosets, we have • H={0, 1}

1H={ 10, 11} ={ 1, 3}.•

2H={ 20, 21} ={ 2, 2}.The partition into right cosets is • H={0, 1}

• H1 ={0 1, 1 1} ={ 1, 2}.• H2={0 2, 1 2} ={ 2, 3}.

Since S3 is not abelian, the partition into left cosets of H is different from the partition into right cosets.

0

0

01

11

122

2

22

0

0 1

0

0

0

1

1

1

2

2

2

1

1

1

1

1

1

1

2

2

2

2

2 2

2

2

3

3

3

3

3 3

3

3

1

The Theorem of Lagrange

Fact: Every coset (left or right) of a subgroup H of a group G has the same number of elements as H.

Theorem of Lagrange

Let H be a subgroup of a finite group G. Then the order of H is a divisor of the order of G.

Proof: Let n be the order of G, and let H have order m. The fact above shows every coset of H also has m elements. Let r be the number of cells in the partition of G into left cosets of H. Then n=rm, so m is indeed a divisor of n.

More Theorem

Corollary

Every group of prime order is cyclic.

Proof: exercise.

Theorem

The order of an element of a finite group divides the order of the group.

Proof: it follows directly from the Theorem of Lagrange.

Index

Definition

Let H be a subgroup of a group G. The number of left cosets of H in G is the index (G: H) of H in G.

Note: if G is finite, then (G: H) is finite and (G: H)=|G| / |H|, since every coset of H contains |H| elements.

Theorem

Suppose H and K are subgroups of a group G such that K H G, and suppose (H : K) and (G: H) are both finite. Then (G: K) is finite, and (G: K)= (G: H) (H : K)