Problem

Let \(H\) be a group acting on a set \(A\). Prove that the relation \(\sim\) on \(A\) defined by $$ a \sim b \quad\text{if and only if}\quad a = hb \quad\text{for some } h \in H $$ is an equivalence relation. (For each \(x \in A\), the equivalence class of \(x\) under \(\sim\) is called the orbit of \(x\) under the action of \(H\). The orbits under the action of \(H\) partition the set \(A\).)

Solution

By definition of group action, \(a = 1a\) for all \(a \in A\). Therefore, \(\sim\) is reflexive.

Suppose \(a \sim b\) for arbitrary \(a,b \in A\). Then by definition of \(\sim\), \(a = hb\) for some \(h \in H\). It follows that

$$ h^{-1}a = h^{-1}(hb) = 1b = b. $$

Therefore, \(b \sim a\) as well. That is, \(\sim\) is symmetric.

Suppose \(a \sim b\) and \(b \sim c\) for arbitrary \(a,b,c \in A\). Then by definition of \(\sim\), \(a = h_1b\) and \(b = h_2c\) for some \(h_1, h_2 \in H\). It follows that

$$ a = h_1b = h_1(h_2c) = (h_1h_2)c $$

Therefore, \(a \sim c\) as well. That is, \(\sim\) is transitive.

Hence, by definition of equivalence relation, \(\sim\) is an equivalence relation.

$$\tag*{$\blacksquare$}$$