Let \(G\) 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 \(\mathrm{Orb}_G(x)\), since \(e(x) = x\) where \(e\) is the identity of \(G\). (This follows by the definition of an action.)

Suppose \(x\) lies in both \(\mathrm{Orb}_G(a)\) and \(\mathrm{Orb}_G(b)\), 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 \(\mathrm{Orb}_G(a) = \mathrm{Orb}_G(b)\); it is an exercise to prove this formally.

Conversely, if \(r \in \mathrm{Orb}_G(a)\), then \(r = m(b)\), say, some \(m \in G\). Then \(r = m(g^{-1}h(b)) = m g^{-1} h (b)\), so \(r \in \mathrm{Orb}_G(b)\). <div><div>