Counter example to the claim that orbits of $S$ constitute a partition of $S$. 
Consider the set $S=\{ (1),(12),(13),(23),(123),(132)\}$. Let $G$ be a group of permutations of a set $S$. Prove that the orbits of the members of $S$ constitute a partition of $S$.

Now
\begin{eqnarray*}
\operatorname{orb}(1)&=&\{ 2,3\}\\ 
\operatorname{orb}(2)&=&\{ 1,3\}\\ 
\operatorname{orb}(3)&=&\{ 1,2\}\\ 
\end{eqnarray*}
These sets are not disjoint. Hence my doubt.
Then how to proceed to prove second statement.
Any hint or suggestion will be appreciated.
 A: Let $e\in G$ denote the identity element and let $x\in S$. Then by definition of an action you have $e\cdot x=x$, so $x\in\operatorname{orb}(x)$. This is where your supposed counterexample fails. Intuitively speaking; every element can be reached from itself by the action of something in $G$; by doing nothing.
A key step in proving that orbits are disjoint is proving that
$$x\in\operatorname{orb}(y)\quad\Leftrightarrow\quad y\in\operatorname{orb}(x).$$
From this it is not hard to show that the relation $\sim$ defined by 
$$x\sim y\quad\Leftrightarrow\ x\in\operatorname{orb}(y),$$
is an equivalence relation. The equivalence classes partition $S$, and these are precisely the orbits.
A: An alternative is to consider the map $orb: S \to \mathcal P(S)$ given by $x \mapsto orb(x)$.
Then the relation $x \sim y$ iff $orb(x)=orb(y)$ is clearly an equivalence relation on $S$, because equality is an equivalence relation.
Therefore, the equivalence classes of this relation form a partition of $S$. These equivalence classes are exactly the orbits. This is the only point that needs a little work.
A: In your example you always forgot that $i \in \text{orb}(i)$ since $(1)(i)=i$ for $i \in {1,2,3}$. Thus all the orbits coincide, and you do not have a counterexample.
For the proof of fact 2): consider two orbits not disjoints, so $a \in \text{orb}(x) \cap \text{orb}(y)$. By definition we have $g,h \in G$ such that $gx=a=hy$, from this you obtain $y=h^{-1}gx$ and now you can easily deduce that the two orbits coincide.
