Group orbits partition

by Patrick Stevens Jun 20 2016

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$~$.


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)$~$. %%