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:

Reflexivity: every $x \in X$ is related to itself, that is, $x \sim x$ .
Symmetry: If $x \sim y$ , then also $y \sim x$ .
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!

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 $x = xe$ .
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 $h^{-1}$ .
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

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

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