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?


About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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