TITLE: Find orbit of $1$ for $\sigma$
QUESTION [7 upvotes]: $\sigma = \left( \begin{array}{cc}1&2&3&4&5&6\\3&1&4&5&6&2\end{array}\right)$
$ 1 \mathop{\rightarrow}^{\sigma} 3 \mathop{\rightarrow}^{\sigma} 4 \mathop{\rightarrow}^{\sigma} 5 \mathop{\rightarrow}^{\sigma} 6 \mathop{\rightarrow}^{\sigma} 2 \mathop{\rightarrow}^{\sigma} 1 \mathop{\rightarrow}^{\sigma} \ldots$ following the notation in my book.
But the question asks the orbit of $1$, so is that the orbit containing one, or one orbit? If it is the orbit containing one it would be $\{1,3,4,5,6,2\}$ 
If that is not what is meant then I am not sure what to do.

REPLY [3 votes]: As the book says, the orbit of $1$ under $\sigma$ is the set $$\{\sigma^n(1) \mid n\in \Bbb Z\}$$
so you just need to look at $$\mathcal O_{1,\sigma} = \{\sigma^0(1) = 1, \sigma(1), \sigma^2(1), \ldots\}$$
and since, as you observed, $\sigma$ maps $$ 1 
\mapsto 3
\mapsto 4
\mapsto 5
\mapsto 6
\mapsto 2
\mapsto 1$$ the orbit of $1$ is the set containing all of these, which is $$\{1,2,3,4,5,6\}.$$
In short, your answer is correct.