# Group orbits partition

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.

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

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>

Parents:

• Group action

“Groups, as men, will be known by their actions.”