# Orbit of Permutations

Let $$f$$ be a permutation on a set $$S$$. If a relation $$ \sim $$ is defined on $$S$$ such that

\[a \sim b \Leftrightarrow {f^{\left( n \right)}}\left( a \right) = b\]

for some integrals $$n\forall \,a,b \in S$$, we observe that the relation is:

**(i) Reflexive:** the relation is reflexive, i.e. $$a \sim a$$, now we can define the reflexive property according to the above definition, because

\[a \sim a \Leftrightarrow {f^{\left( n \right)}}\left( a \right)\]

\[ = I\left( a \right) = a,\forall a \in S\]

**(ii) Symmetric:** the relation is symmetric, i.e. $$a \sim b \Rightarrow b \sim a$$, we can show this relation by using the definition of orbit of permutation, because

\[a \sim a \Leftrightarrow {f^{\left( n \right)}}\left( a \right) = b\]

for some integers $$n$$

\[ \Rightarrow a = {f^{ – \left( n \right)}}\left( b \right)\]

\[ \Rightarrow b \sim a,\,\forall a,b \in S\]

**(iii) Transitive:** the above relation is transitive, i.e. $$a \sim b$$ and $$b \sim c$$ implies $$a \sim c$$, now we can prove this transitive property by using the above definition of orbit of permutation, because

\[ \Rightarrow {f^{\left( n \right)}}\left( a \right) = b,\,\,\,{f^{\left( m \right)}}\left( b \right) = c\] for some integers $$n$$ and $$m$$

\[ \Rightarrow {f^m}\left( {{f^n}\left( a \right)} \right) = {f^m}\left( b \right) = c\]

\[ \Rightarrow {f^{m + n}}\left( a \right) = c\] for some integer $$m + n$$

\[ \Rightarrow a \sim c\]

Thus the above defined relation $$ \sim $$ is an equivalence relation on $$S$$ and hence we partition it into mutually disjoint classes. Each equivalence class determined by the relation is called an orbit of $$f$$.