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.”