Let be a group, acting on the set $X$ . Then the orbits of $X$ under $G$ form a partition of $X$ .

Proof

We need to show that every element of $X$ is in an orbit, and that if $x \in X$ lies in two orbits then they are the same orbit.

Certainly $x \in X$ lies in an orbit: it lies in the orbit ${O r b}_{G} (x)$ , since $e (x) = x$ where $e$ is the identity of $G$ . (This follows by the definition of an action.)

Suppose $x$ lies in both ${O r b}_{G} (a)$ and ${O r b}_{G} (b)$ , where $a, b \in X$ . Then $g (a) = h (b) = x$ for some $g, h \in G$ . This tells us that $h^{- 1} g (a) = b$ , so in fact ${O r b}_{G} (a) = {O r b}_{G} (b)$ ; it is an exercise to prove this formally.

Show solution

Indeed, if $r \in {O r b}_{G} (b)$ , then $r = k (b)$ , say, some $k \in G$ . Then $r = k (h^{- 1} g (a)) = k h^{- 1} g (a)$ , so $r \in {O r b}_{G} (a)$ .

Conversely, if $r \in {O r b}_{G} (a)$ , then $r = m (b)$ , say, some $m \in G$ . Then $r = m (g^{- 1} h (b)) = m g^{- 1} h (b)$ , so $r \in {O r b}_{G} (b)$ .