# 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.”