Recall from last time we introduced the notion of a cyclic group, and said that we’d introduce cosets. For this, we need the notion of partitions and equivalence relations (which are handy in more general scenarios). As I said previously, this is just a transcript of a talk given at the abstract algebra seminars. This talk was given by my good friend, Nadir Akhtar. Since I LiveTeXed it, any errors are mine. (I’ve added a proof that equivalence relations are basically the same as partitions at the end.)

Today we’re going to be talking about equivalence relations between different groups. There are three parts to the definition.

**Definition. **X is a set, equivalence relation .

- reflexive: for every , .
- symmetry: for all , if and only if .
- transitive property: for every , if and then .

All three of these conditions have to be met. Some examples: suppose we have a set (examples below), and . Define as . Then this is an example of an equivalence relation. For e.g. , etc. Here’s a non-example. Suppose is one of the above sets; then define if . This isn’t an equivalence relation. Now suppose , and let . Say that if and only if . This is an equivalence relation. Here’s a nice question: is there an example of a relation on , say, which is *not* transitive?

(A quick review of subgroups.) Consider a group and a subgroup .

**Definition. **Define the following equivalence relation on : say that if and only if .

Let’s prove that this is an equvialence relation.

*Proof. *Let . Then . So . Proves reflexivity. Now symmetry. Suppose . Then . If this is true, then it has an inverse which is also in , because is a subgroup. Now, . So , and so . Move on to transitivity. Let . Suppose and . Then . Then multiply them to get . Thus . QED.

So another definition is that of the* equivalence class*. Let’s say that is an equivalence relation on a set , and let . Then the equivalence class of , denoted , is defined as . For example, , and let be . Then . Suppose is . Then if is odd (resp. even) then is the set of odd numbers (resp. even numbers). (some clarification on modular arithmetic.)

Now onto partitions. Let be a set. A partition of is a collection of subsets , with , such that:

- (pizza used as an example here; now I’m hungry).
- For any , then .

Significance of partitions are: an equivalence relation gives rise to a unique partition, and vice-versa. Suppose we’ve got , with given by . Then and are disjoint, and cover .

We’ll prove that *an equivalence relation gives rise to a unique partition, and vice-versa*. Let’s do the first part now. Let be a set, with equivalence relation . Then by reflexivity. Since every element is in some equivalence class, the union of all the equivalence classes contains itself. So we need to show that none of the equivalence classes intersect each other. Now if , then . So let . Then . And since , . But also by symmetry, so if then .

Now let and , and let . Since , . Since , . Because of the above, therefore, ! Thus every equivalence class gives rise to a unique partition.

Let be a partition of , and define on by if and only if . It’s easy to check that this is an equivalence relation. Reversing the above argument shows that *an equivalence relation gives rise to a unique partition, and vice-versa*!

This is really quite cool. Let , considered as a group under addition. Let be the equivalence relation on a group with a chosen subgroup earlier, where the subgroup here is . Then this gives rise to an equivalence relation, namely that if and only if is a multiple of , i.e., if and only if ; but then !

Similarly, we can generalize this to *all* groups, which leads to the notion of a coset. Let be a group and let . Let be the equivalence relation defined on a group with a chosen subgroup defined earlier, and let . Let denote the equivalence class of under this equivalence relation. Then . This is called the **left coset** of .

To see that , suppose . Then . Write ; then . Since , we see that . Now let . Then for some ; but then , so . Then , so . QED.

Next time we’ll prove Lagrange’s theorem, which, unfortunately, has had to be postponed yet again!

Have fun,

SKD