Orbits are the minimal $G$-invariant subsets A subset $\phi \subseteq \Omega $ is G-invariant ($G \le$ Sym$(\Omega)$ ) if $(\forall \sigma \in G)(\phi^\sigma = \phi)$. The orbits of the action are the equivalence classes of the relation $\Omega$ defined as " $x \equiv_G y$" if $(\exists\sigma \in G)(x^\sigma = y)$.
My Question : Orbits are the minimal $G$-invariant subsets. I know that orbits are the partitions of the set $\Omega$ but not able to prove that "Orbits are the minimal $G$-invariant subsets".
 A: Suppose that $G$ acts on $X$ and $x$ an element of $X$, let $Y$ a subset of the orbit of $x$ which is invariant by $G$, let $y\in Y$, $y=g.x$ implies $x=g^{-1}.y$ we deduce $x\in Y$ and $Y=X$.
A: Here is a roadmap:


*

*An orbit is a $G$-invariant subset

*A $G$-invariant subset is a union of orbits

*If a $G$-invariant subset contains more than one orbit, then it is not minimal.

*A minimal $G$-invariant subset is an orbit
A: Think about it like this. What's an orbit? $G$ is some random group that acts on some set $\Omega$. Take any element $x$ in $\Omega$. Every element $g$ of $G$ sends $x$ to some other element of $\Omega$ -- call that point $gx$. The set of elements of the form $gx$ is called the orbit of $x$. It's just what you get when you take every element of $G$ and see where it sends $x$. Any orbit is $G$-invariant. Furthermore, it is a minimal $G$-invariant set because if you took some proper subset of it, you're "missing" some of the points that $x$ gets sent to. In the hint above I gave a suggestion as to how to give a proof.
Here's a proof. Let $O$ be an orbit of $G$. Let $x \in O$. Then $O=Gx$. We claim that $O$ is a minimal $G$-invariant subset. Let $H$ be a proper subset of $O$. We will show that $H$ is not $G$-invariant. Let $y$ be such that $y \in O$ but $y \notin H$. Then $y = gx$ for some $g \in G$. We claim that there exists an element $x' \in H$ such that for some $g' \in G,$ $g'x' \notin H$. In fact, we can take $x'$ to be any element $h \in H$. Then $h$ has the form $g''x$ for some $g \in G$. Then $g(g'')^{-1}h=gx=y$, which is not in $H$. Therefore $H$ is not invariant under $G$. 
