# Group Theory/Cosets and Lagrange's theorem

**Definition (left coset)**:

Let be a group, and a subgroup, and . Then the **left coset** of represented by is the set

- .

Right cosets are defined in an analogous fashion:

**Definition (right coset)**:

Let be a group, and a subgroup, and . Then the **right coset** of represented by is the set

- .

For both of these, we have the following proposition:

**Proposition (being in the same left coset is an equivalence relation)**:

Let be a group, and define a relation on by

- .

Then is an equivalence relation, and we also have the formula

- .

**Proof:** We first prove the alternative formula for being in the same coset. If and , we find from the latter equation an so that , so that and hence , the latter identity because is a group and in particular closed under inversion. On the other hand, if , then for some , and hence (because the identity is in ) and then .

Hence, the two formulae for the relations coincide, and it remains to check that we're dealing with an equivalence relation. Indeed, suppose and . Then there exist so that and so that . Then , so that , ie. , proving transitivity. Reflexivity follows since the identity is in , and symmetry follows because implies and hence .

Analogously, we have the following proposition:

**Proposition (being in the same right coset is an equivalence relation)**:

Let be a group, and define a relation on by

- .

Then is an equivalence relation, and we also have the formula

- .

**Proof:** Consider the opposite group of . There is a bijection , given by , under which two elements belonging to the same right coset of correspond to elements belonging to the same *left* coset of . But the relation defined by the latter was seen to be an equivalence relation.

**Definition (index)**:

Let be a group, and let be a subgroup. Then the **index** of is defined to be the number

- .

That is, the index is precisely the number of left cosets.

**Proposition (Lagrange's theorem)**:

Let be a group, and let be a finite subgroup. Then

- .

In particular, the order of divides the order of .

**Proof:** We have seen that being in the same left coset is an equivalence relation, so that the equivalence classes partition . Moreover, every equivalence class (ie. coset) has the same cardinality as via the bijection .

**Proposition (number of right cosets equals number of left cosets)**:

Let be a group, and a subgroup. Then the number of right cosets of equals the number of left cosets of .

**Proof:** By Lagrange's theorem, the number of left cosets equals . But we may consider the opposite group of . Its left cosets are almost exactly the right cosets of ; only the orders of the products are interchanged. But in particular, the number of left cosets of , which, by Lagrange's theorem equals , is equal to , which is what we wanted to prove.

Hence, we may also use the notation for the number of *right* cosets.

**Proposition (degree formula)**:

Let be a group, and let . Then

- .

**Proof:** We may partition into a family of left cosets , where for all we have . Moreover, may be partitioned into a family of left cosets of . Then is a family of left cosets of that partitions (since each element is in one left coset of , and then is in a unique coset , and then is the unique coset in which is), and the cardinality of this family, which is , is the number of left cosets of in .

## Exercises

edit- Prove that , thus establishing another formula for the equivalence relation of being in the same coset.
- Formulate Lagrange's theorem for right cosets, without using index notation.