MaBloWriMo 27: From subgroups to equivalence relations

Again, let G be a group and H a subgroup of G. Then we can define a binary relation on elements of G, called \sim_H, as follows:

x \sim_H y if and only if there is some h \in H such that x = yh.

That is, for any two elements x, y \in G, either x \sim_H y, or not: yes, if you can get from y to x by combining (on the right) with some element in H, and otherwise, no. Note that given any two elements x, y \in G, it is always possible to get from y to x by combining with some element of G: in particular, x = y(y^{-1}x). But this might not be an element of H.

Now, an equivalence relation on a set X is a relation x \sim y with the following three properties:

  1. Reflexivity: every x \in X is related to itself, that is, x \sim x.
  2. Symmetry: If x \sim y, then also y \sim x.
  3. Transitivity: If x \sim y and y \sim z, then x \sim z.

The usual equality relation satisfies these properties: things are always equal to themselves; if x = y then y = x; and if x = y and y = z then x = z. The notion of an equivalence relation is a way to talk about more general kinds of equality.

Let’s prove that \sim_H is an equivalence relation. This is really cool because it turns out that the three properties of an equivalence relation each follow from one of the three properties of a group!

  1. Reflexivity. We have to show that any x \in G is related to itself, that is, x \sim_H x. By definition this means there is some h \in H such that x = xh. Well, that’s easy: since H is a group, it has to contain the identity element e, and x = xe.
  2. Symmetry. Suppose x \sim_H y, that is, x = yh for some h  \in H. Then we have to show y \sim_H x, that is, there is some h' \in H (which could be different from h) such that y =  xh'. Well, since H is a group, it has to contain inverses. We can combine both sides of x = yh with h^{-1} to obtain xh^{-1}  = y—so the h' we are looking for is precisely h^{-1}.
  3. Transitivity. Suppose x \sim_H y and y \sim_H z. That means there are h_1, h_2 \in H for which x = y h_1 and y = z h_2. We want to show that x \sim_H z, that is, x = zh for some h  \in H. Substituting for y, we find that x = y h_1 = (z h_2) h_1 = z (h_2 h_1) (note how we used the third property of a group, associativity). So the h we are looking for is just h_2 h_1, which has to be in H since H is closed under the binary operation.

So for a given subgroup H \leq G, this relation defines a sort of “equality with respect to H” on the elements of G (whatever that means!). As for an example—consider again the subgroup \{0,4\} \leq \mathbb{Z}_8. Which elements of \mathbb{Z}_8 are related to each other under \sim_{\{0,4\}}? What do you notice?

Advertisements

About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in algebra, group theory, proof and tagged , , , , , , . Bookmark the permalink.