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**Formal Definition**

Define an equivalence relation $\sim$ such that, for $a,b\in A$, $a\sim b$ if and only if $b=\sigma ^n(a)$ for some $n\in \mathbb{Z}$ and permutation $\sigma$ on $A$.

Let $\sigma$ be a permutation on a set $A$. Then the equivalence classes in $A$ determined by the equivalence relation $\sim$ are the **orbits** of $\sigma$.

**Informal Definition**

An **orbit** is a certain type of set that is associated with a permutation $\sigma$ of a set $A$ that is defined as follows:

If you start out with an element of $A$ and repeatedly apply $\sigma$ to it, then the set of all the elements you obtain is an **orbit** of $\sigma$.

**Example(s)**

Consider the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 & 8 & 6 & 7 & 4 & 1 & 5 & 2 \end{pmatrix}$ in $S_8$

To find the orbit containing $1$ we apply $\sigma$ repeatedly, obtaining

(1)Since $\sigma^{-1}$ would simply reverse the directions of the arrows in the chain, we see that the **orbit** containing $1$ is $\{1,3,6\}$. Choosing other integers in the set permuted by $\sigma$, we perform the same procedure until all elements mapped by $\sigma$ have been placed in an **orbit**. In this way, the total list of the orbits of $\sigma$ is found to be

**Non-example(s)**

Consider the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \end{pmatrix}$ in $S_8$.

Then $(3\ 4)$ is NOT an **orbit** of $\sigma$, since it is not a set. ( $(3\ 4)$ is a cycle )

Also, $\{1, 2\}$ is NOT and **orbit** of $\sigma$, since repeated application of $\sigma$ to $1$ will never yield $2$.

**Additional Comments**

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