# Group orbits partition

When a group acts on a set, the set falls naturally into distinct pieces, where the group action only permutes elements within any given piece, not between them.

Let $G$ be a Group, acting on the set $X$. Then the orbits of $X$ under $G$ form a [set_partition 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.

%%hidden(Show solution): 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 $r \in \mathrm{Orb}_G(b)$. %%