If H is a subgroup of G, you can break G up into pieces, each of which looks like H:

These pieces are called * cosets* of H, and they
arise by "multiplying" H by elements of G.

* Definition.* Let G be a group and let .
A * left coset* of H in G is a subset of the form

The element g is a * representative* of the coset
. The collection of left cosets is denoted .

Likewise, a * right coset* is a subset of the
form

The set of right cosets is denoted .

Thus, the left coset consists of g times everything in H; consists of everything in H times g.

I've written everything as if the operation in the group was "multiplication". The case when the operation is "addition" is discussed in an example below.

* Example.* (* Listing the
elements of cosets*) (a) List the elements of
and the elements of the cyclic subgroup generated by 9.

(b) List the elements of the cosets of in .

(a)

(b) The subgroup is always a coset. I'll list that first:

Take an element of which is not in the subgroup --- say 3. Multiply the subgroup by the element:

Take an element of which is not in either of the two known cosets --- say 5. Multiply the subgroup by the element:

Notice that all the cosets have 3 elements --- the same as the number of elements in the subgroup.

At this point, there are only 3 elements which aren't in any of the known cosets. These elements make up the last coset: . You can check that

3 represents the coset , but a given coset can
be represented by *any* of its elements. For example,

* Example.* (* Listing the
elements of cosets*) List the elements of the cosets of in .

consists of two cosets: the even numbers and the odd numbers. Explicitly,

Notice that when the operation in the group is +, a coset of a subgroup H is written .

* Example.* (* Listing the
elements of cosets*) List the elements of the cosets of the
subgroup of the group of quaternions.

Here is the table for the group of quaternions:

Consider the subgroup . Its cosets are

There are four distinct cosets. Notice that . This is a
special case of * Lagrange's theorem*: The order
of a subgroup times the number of cosets of the subgroup equals the
order of the group.

* Example.* (* Identifying a set
of cosets with another set*) Show that the set of cosets can be identified with , the group of complex
numbers of modulus 1 under complex multiplication.

The cosets are

Thus, there is one coset for each number in the half-open interval .

On the other hand, you can "wrap" the half-open interval around the circle in the complex plane: Use , . It's easy to show this is a bijection by constructing an inverse using the logarithm.

Thus, there is a bijection from the set of cosets to the circle .

In fact, this is an example of an * isomorphism*
of groups.

* Theorem.* Let G be a group and let .
The left cosets of H in G form a partition of G.

* Proof.* I need to show that the union of the
left cosets is the whole group, and that different cosets do not
overlap.

Let . Since , it follows that is in . This shows that every element of G lies in some coset of H, so the union of the cosets is all of G.

Next, suppose and are two cosets of H, and suppose they are not disjoint. I must show they're identical: . As usual, I can show two sets are equal by showing that each is contained in the other.

Since and are not disjoint, I can find an element . Write for . Then

Now let . Then

The element on the right is in , since it is b times something in H. Therefore, , and . By symmetry, , so .

* Theorem.* Any two left cosets have the same
number of elements.

* Proof.* Let H be a subgroup of a group G, and
let . I must show that and have the same number
of elements. *By definition*, this means that I must construct
a bijective map from to .

An element of looks like , for some . So it is tempting to simply define by

But how do you know this is *well-defined*? How do you know
that the *same element* of might not be expressed as both
and , where h and are *different*
elements of H?

Fortunately, this can't happen; if , then

Thus, it's legitimate for me to define a function f as above.

Likewise, I can define by

This is well-defined, just as f was.

Since f and g are clearly inverses, f (or g) is a bijection, and and have the same number of elements.

* Definition.* If G is a group and ,
is called the * index* of H in G,
and is denoted .

The way I've defined it, the index of H in G is the number of
*left* cosets of H. It turns out that this is the same as the
number of right cosets.

* Theorem.* (* Lagrange's
theorem*) Let G be a finite group and let H be a subgroup of G.
Then

* Proof.* The cosets of H partition G into
pieces, and each piece contains elements. So the total number of
elements in the pieces is , but this
is all of G:

Now divide both sides by .

Note that this result implies that *the order of a subgroup
divides the order of the group*. Thus, a group of order 14
*could* have subgroups of order 1, 2, 7, or 14, but *could
not* have a subgroup of order 5.

* Example.* (* A specific example
of Lagrange's theorem*) Verify Lagrange's theorem for the
subgroup of .

The cosets are

Notice there are 3 cosets, each containing 2 elements, and that the cosets form a partition of the group.

* Example.* (* A specific example
of Lagrange's theorem*) List the elements of the cosets of in .

First, list the elements of the subgroup:

The subgroup is a coset.

The subgroup has 6 elements and the group has 24. By Lagrange's theorem, there are 4 cosets.

isn't in the subgroup; add it to the subgroup:

isn't in either of the known cosets; add it to the subgroup:

The remaining elements make up the fourth coset. I can find them by noting that isn't in the three known cosets, so the fourth coset is represented by :

Notice that there are 4 cosets, each containing 6 elements, and the cosets form a partition of the group.

* Corollary.* Every group of prime order is
cyclic.

* Proof.* Suppose G is a group of order p, where
p is prime. Let , . is a
subgroup of G, and since ,
.

But divides by Lagrange's theorem, and the only positive numbers which divide are 1 and p. Therefore, , which means that is all of G. That is, G is cyclic with generator g.

For example, this means that the only group of order 17 is the cyclic group of order 17.

I noted earlier that the number of left cosets equals the number of right cosets; here's the proof.

* Proposition.* Let G be a group,
. The set of left cosets may be put in 1-1 correspondence
with the set of right cosets .

* Proof.* Define by . I need to show
is well-defined.

Suppose . Then , so for some . Then

Next, define by . A computation similar to the one I just did shows is well-defined. and are inverses, so either one gives a bijection of with .

While there are the same *number* of left and right cosets,
the left and right cosets may be different as sets. In fact, if the
left and right cosets are the same as sets, the subgroup is said to
be * normal*. It's a very important condition on
a subgroup, since it will allow us to turn the set of left (or right)
cosets into a * quotient group*.

* Example.* (* A subgroup whose
left and right cosets are different*) List the elements of the
left cosets and the right cosets of the subgroup of .

The left cosets are

The right cosets are

The left and right cosets aren't the same, though there are the same number of left and right cosets.

Copyright 2018 by Bruce Ikenaga