# Double cosets

We turn now to another important decomposition of an arbitrary group G into disjoint complexes. Such decompositions will play an important role in our later considerations of the Sylow Theorems.

Let be an arbitrary group and let and be subgroups of . For , we define

 (8.3)

where and . We contend first of all that the relation given in (8.3) is an equivalence relation on . To see this, we note

1. (Reflexivity): since , where is the identity element and and since , .

2. (Symmetry): implies there exist and such that , but then , so since and .

3. (Transitivity): If and , then there exist elements and such that

whence

Since and , we have that .

Thus given by (8.3) is indeed an equivalence relation on . We next take a look at the equivalence classes.

Definition 8.2.1   Let be a group with subgroups and (not necessarily distinct). If , the complex is called a double coset with respect to and . By definition,

For , the equivalence class of as we recall, contains all with . By the definition given in (8.3), this means , where and . Thus . As the above statements are all if and only if'', we see that , the double coset given in Definition 8.2.1. By our general theorem on equivalence relations, Theorem 1.1.4, we know that either

and

where the union is taken over certain . The identity element belongs to the complex .

If , then we simply get the right coset decomposition of with respect to . If , then we have the left coset decomposition of with respect to . Thus the double coset decomposition of a group may be viewed as a generalization of the coset (right or left) decomposition of a group. However, the reader should be careful not to generalize all facts related to coset decompositions to the case of double coset decompositions. For example, we saw that any two cosets of a finite group have the same number of elements. We shall presently see that this is not the case with double cosets.

Let us consider the double coset . Clearly contains all right cosets of the form , where and contains all left cosets of the form where . We claim, as a matter of fact, that is a union of right or left cosets of the above form. For suppose that

Then there exist elements and such that

or . This implies that

and so . Since this shows that any left coset which has anything at all in common with , must be totally contained in , we have
 (8.4)

Similarly, it can be shown that
 (8.5)

Next, we wish to ascertain the number of left and right cosets in the double coset. Even though this number can be finite for an infinite double coset, we assume . This is contained in

Theorem 8.2.2   Let be a finite group, let , , and let . Then
(a)
The number of right cosets of in is .

(b)
The number of left cosets of in is .

Proof: We first note that is a subgroup by Proposition 7.2.1. Consider the mapping of the double coset onto the complex given by . It is easy to show that this map is well-defined, 1-1, and onto (see exercise 2 for this section). Thus . But is the product of two subgroups and so by the product theorem (Theorem 4.3.6),

Now according to (8.4), the number of left cosets of in is . Thus the number of left cosets of in is

This establishes part (b) of the theorem. A similar argument establishes part (a) (this is left as an exercise).

Under the same hypotheses as in Theorem 8.2.2, we use the notation to be the number of right cosets of in times (note from (8.5) that ). Thus Theorem 8.2.2 implies that

This proves the following result.

Corollary 8.2.3   Let be a finite group and let , . If (disjoint), then
 (8.6)

where .

Subsections

David Joyner 2007-08-06