## 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?